A Line Parallel to One Side of a Triangle Cuts the Other Two in the Same Ratio
Separate congruence from similarity, state the two conditions two polygons must satisfy to be similar, prove the Basic Proportionality Theorem, use it and its converse to find unknown lengths and to test whether a line is parallel, and apply it inside a trapezium.
What is the difference between two figures that match and two that merely match in shape?
Hold two photographs of the same building side by side, one a postcard and one a poster. Every angle in the two pictures is identical and every length in the poster is the same multiple of the matching length in the postcard. The pictures are not the same size, so they are not congruent — but they are unmistakably the same shape.
That relationship is called similarity, and it is what this chapter is about.
- Congruent figures have the same shape and the same size. One can be placed exactly on top of the other
- Similar figures have the same shape but not necessarily the same size. One is an enlargement of the other
**So congruence is the special case of similarity where the enlargement factor is . Every pair of congruent figures is similar; most similar figures are not congruent.
And then the chapter's central result appears, which is more useful than it first looks. Draw any line parallel to one side of a triangle, cutting the other two sides. That line divides those two sides in exactly the same ratio — and it does so no matter where you draw it, how long the sides are, or what shape the triangle has.
That statement is the Basic Proportionality Theorem, and with its converse it does three jobs:
- Find an unknown length in a triangle crossed by a parallel line
- Prove that a line is parallel to a side, by checking that two ratios are equal
- Establish ratios inside a trapezium**, where two parallel sides already exist
This page covers the first part of the CBSE Class 10 Maths chapter on triangles: congruence and similarity, the Basic Proportionality Theorem and its proof, the converse, and applications in triangles and trapeziums.
That relationship is called similarity, and it is what this chapter is about.
- Congruent figures have the same shape and the same size. One can be placed exactly on top of the other
- Similar figures have the same shape but not necessarily the same size. One is an enlargement of the other
**So congruence is the special case of similarity where the enlargement factor is . Every pair of congruent figures is similar; most similar figures are not congruent.
And then the chapter's central result appears, which is more useful than it first looks. Draw any line parallel to one side of a triangle, cutting the other two sides. That line divides those two sides in exactly the same ratio — and it does so no matter where you draw it, how long the sides are, or what shape the triangle has.
That statement is the Basic Proportionality Theorem, and with its converse it does three jobs:
- Find an unknown length in a triangle crossed by a parallel line
- Prove that a line is parallel to a side, by checking that two ratios are equal
- Establish ratios inside a trapezium**, where two parallel sides already exist
This page covers the first part of the CBSE Class 10 Maths chapter on triangles: congruence and similarity, the Basic Proportionality Theorem and its proof, the converse, and applications in triangles and trapeziums.
What exactly must be true for two polygons to be similar?
Two conditions, and both are needed: corresponding angles must be equal, and corresponding sides must be in the same ratio.
Written out for two polygons with the same number of sides:
- Their corresponding angles are equal
- Their corresponding sides are in the same ratio, that is, proportional
The common value of those ratios is the scale factor of the enlargement.
Why one condition alone is not enough, and this is the most examined idea in the section.
Equal angles without proportional sides. Take a cm by cm rectangle and a cm by cm rectangle. Every angle in both is a right angle, so all corresponding angles are equal. But the side ratios are and , which are different. The two rectangles are not similar, and no amount of enlarging one will turn it into the other.
Proportional sides without equal angles. Take a square of side cm and a rhombus of side cm whose angles are and . **All four ratios of corresponding sides equal **, so the sides are proportional. But a right angle is not a angle. The square and the rhombus are not similar.
Those two examples together are the standard answer to "why are both conditions necessary?", and an examiner expects a counterexample, not just an assertion.
Which figures are always similar, whatever their size?
- All circles — a circle has no angles to go wrong and one length to scale
- All squares — every angle is and all four sides are equal, so every ratio is the same
- All equilateral triangles — every angle is and all three sides are equal
And which are not?
- Two rectangles need not be similar, as shown above
- Two rhombuses need not be similar
- Two isosceles triangles need not be similar
The special thing about triangles, which the next part of the chapter proves. For polygons in general you must check both conditions. For triangles, either one on its own forces the other — if the angles of two triangles match, their sides are automatically proportional, and if the sides are proportional, the angles automatically match. Triangles are the only polygons with that property, and it is what makes them the tool for every practical measurement in this chapter.
One point of notation that carries marks. Writing asserts a correspondence: with , with , with . So it claims that and that
The order of the letters is part of the statement. Writing when the correspondence is makes every ratio that follows wrong, and it is the commonest way to lose marks in a similarity proof.
Written out for two polygons with the same number of sides:
- Their corresponding angles are equal
- Their corresponding sides are in the same ratio, that is, proportional
The common value of those ratios is the scale factor of the enlargement.
Why one condition alone is not enough, and this is the most examined idea in the section.
Equal angles without proportional sides. Take a cm by cm rectangle and a cm by cm rectangle. Every angle in both is a right angle, so all corresponding angles are equal. But the side ratios are and , which are different. The two rectangles are not similar, and no amount of enlarging one will turn it into the other.
Proportional sides without equal angles. Take a square of side cm and a rhombus of side cm whose angles are and . **All four ratios of corresponding sides equal **, so the sides are proportional. But a right angle is not a angle. The square and the rhombus are not similar.
Those two examples together are the standard answer to "why are both conditions necessary?", and an examiner expects a counterexample, not just an assertion.
Which figures are always similar, whatever their size?
- All circles — a circle has no angles to go wrong and one length to scale
- All squares — every angle is and all four sides are equal, so every ratio is the same
- All equilateral triangles — every angle is and all three sides are equal
And which are not?
- Two rectangles need not be similar, as shown above
- Two rhombuses need not be similar
- Two isosceles triangles need not be similar
The special thing about triangles, which the next part of the chapter proves. For polygons in general you must check both conditions. For triangles, either one on its own forces the other — if the angles of two triangles match, their sides are automatically proportional, and if the sides are proportional, the angles automatically match. Triangles are the only polygons with that property, and it is what makes them the tool for every practical measurement in this chapter.
One point of notation that carries marks. Writing asserts a correspondence: with , with , with . So it claims that and that
The order of the letters is part of the statement. Writing when the correspondence is makes every ratio that follows wrong, and it is the commonest way to lose marks in a similarity proof.
Formula
How do you prove the Basic Proportionality Theorem?
By comparing areas. Two triangles on the same base and between the same parallels have equal areas, and two triangles with the same height have areas in the ratio of their bases.
The theorem states: if a line is drawn parallel to one side of a triangle, intersecting the other two sides in distinct points, then it divides those two sides in the same ratio.
In with on and on and ,
The proof, step by step. Join and , and drop perpendiculars onto and onto .
**Step 1 — two triangles with the same height along .** Triangles and both have their bases on the line and share the vertex , so they have the same height :
**Step 2 — two triangles with the same height along .** Triangles and both have their bases on the line and share the vertex , so they have the same height :
Step 3 — the parallel line enters. Triangles and stand on the same base and lie between the same parallels and . Two triangles on the same base and between the same parallels have equal areas, so
Step 4 — combine. The two ratios from Steps 1 and 2 have the same numerator and now equal denominators, so they are equal:
which proves the theorem.
Notice exactly where the parallel condition was used — only in Step 3, to make the two areas equal. That is the whole content of the hypothesis, and a proof that does not point to it is incomplete.
The equivalent forms you are allowed to quote. Starting from , adding to both sides gives
and taking reciprocals of the original and adding gives
All three are the same theorem, and choosing the right one for the lengths a question actually gives you saves a rearrangement. **If the question gives you and , use the third form directly rather than computing first.
The converse, which is a separate result and must be quoted by name. If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.** The theorem lets you find lengths; the converse lets you prove parallelism, and questions ask for each about equally often.
The theorem states: if a line is drawn parallel to one side of a triangle, intersecting the other two sides in distinct points, then it divides those two sides in the same ratio.
In with on and on and ,
The proof, step by step. Join and , and drop perpendiculars onto and onto .
**Step 1 — two triangles with the same height along .** Triangles and both have their bases on the line and share the vertex , so they have the same height :
**Step 2 — two triangles with the same height along .** Triangles and both have their bases on the line and share the vertex , so they have the same height :
Step 3 — the parallel line enters. Triangles and stand on the same base and lie between the same parallels and . Two triangles on the same base and between the same parallels have equal areas, so
Step 4 — combine. The two ratios from Steps 1 and 2 have the same numerator and now equal denominators, so they are equal:
which proves the theorem.
Notice exactly where the parallel condition was used — only in Step 3, to make the two areas equal. That is the whole content of the hypothesis, and a proof that does not point to it is incomplete.
The equivalent forms you are allowed to quote. Starting from , adding to both sides gives
and taking reciprocals of the original and adding gives
All three are the same theorem, and choosing the right one for the lengths a question actually gives you saves a rearrangement. **If the question gives you and , use the third form directly rather than computing first.
The converse, which is a separate result and must be quoted by name. If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.** The theorem lets you find lengths; the converse lets you prove parallelism, and questions ask for each about equally often.
How do you use the theorem and its converse on actual lengths?
For a length, write the theorem in whichever form matches the lengths you are given, substitute, and solve. For parallelism, compute both ratios independently and compare them.
Worked example 1 — a missing length. In , lies on and on with . If cm, cm and cm, find .
Check: and . The ratios match, so cm.
Worked example 2 — an algebraic length. In with , the lengths are , , and . Find .
Cross-multiplying:
Check by substituting back: the lengths become , , and , giving and . Equal, and all four lengths are positive — which is worth confirming, since a value of making any length negative would have to be rejected.
**Notice what happened to the terms. They cancelled, leaving a linear equation. That is typical of this family, and if they do not cancel you should expect a quadratic and the usual rejection of an inadmissible root.
Worked example 3 — the converse, testing for parallelism.** In , the point lies on and on , with cm, cm, cm and cm. Is ?
**The two ratios are equal, so by the converse of the Basic Proportionality Theorem, .
Worked example 4 — the converse, where the answer is no.** In , cm, cm, cm and cm. Is ?
Comparing as decimals, while . **The ratios are unequal, so is not parallel to .
Cross-multiplication is the safer comparison**: and , and . That avoids rounding altogether, and it is the method to use when the two fractions are close.
Worked example 5 — the converse with whole sides given. In , lies on and on with cm, cm, cm and cm. Is ?
First convert the whole sides into the parts the theorem needs:
Then
**Equal, so .
Alternatively use the whole-side form directly**: and . Same conclusion in half the working, because the form was chosen to match the given lengths.
The mistake this example is built to catch. Comparing with — one whole-side ratio against one part ratio. Both fractions must be of the same kind, either part-to-part or part-to-whole, and mixing them gives a wrong verdict on parallelism even when every length was read correctly.
Worked example 1 — a missing length. In , lies on and on with . If cm, cm and cm, find .
Check: and . The ratios match, so cm.
Worked example 2 — an algebraic length. In with , the lengths are , , and . Find .
Cross-multiplying:
Check by substituting back: the lengths become , , and , giving and . Equal, and all four lengths are positive — which is worth confirming, since a value of making any length negative would have to be rejected.
**Notice what happened to the terms. They cancelled, leaving a linear equation. That is typical of this family, and if they do not cancel you should expect a quadratic and the usual rejection of an inadmissible root.
Worked example 3 — the converse, testing for parallelism.** In , the point lies on and on , with cm, cm, cm and cm. Is ?
**The two ratios are equal, so by the converse of the Basic Proportionality Theorem, .
Worked example 4 — the converse, where the answer is no.** In , cm, cm, cm and cm. Is ?
Comparing as decimals, while . **The ratios are unequal, so is not parallel to .
Cross-multiplication is the safer comparison**: and , and . That avoids rounding altogether, and it is the method to use when the two fractions are close.
Worked example 5 — the converse with whole sides given. In , lies on and on with cm, cm, cm and cm. Is ?
First convert the whole sides into the parts the theorem needs:
Then
**Equal, so .
Alternatively use the whole-side form directly**: and . Same conclusion in half the working, because the form was chosen to match the given lengths.
The mistake this example is built to catch. Comparing with — one whole-side ratio against one part ratio. Both fractions must be of the same kind, either part-to-part or part-to-whole, and mixing them gives a wrong verdict on parallelism even when every length was read correctly.
How does the theorem work inside a trapezium and with mid-points?
A trapezium already has a pair of parallel sides, so the theorem applies to any triangle you can find inside it — and the diagonals create exactly such triangles.
Result 1 — the diagonals of a trapezium. In trapezium with , the diagonals meet at . Then
How to prove it. Draw a line through parallel to , meeting at . Since , this line is parallel to both. In , the line is parallel to , so by the theorem
and in , the line is parallel to , so
**Both equal the same ratio , so they equal each other.** That is the whole proof, and the construction line through is the only idea in it.
Worked example — an algebraic version. In trapezium with , the diagonals meet at . Given , , and , find .
Expanding both sides:
Now the rejection, and it is the point of the question. If then , a negative length, which is impossible. **So .
Check with **: , , , , and . Correct, and every length is positive.
Result 2 — mid-points. If and are the mid-points of and in , then
so by the converse of the theorem, .
That is the mid-point theorem, recovered as a special case. The Basic Proportionality Theorem is the general statement and the mid-point theorem is what it says when the ratio happens to be — which is a good example of a familiar result turning out to be a corollary of a stronger one.
Result 3 — a line through one mid-point. If is the mid-point of and the line through parallel to meets at , then
so ** is the mid-point of .** A line through the mid-point of one side, parallel to another side, bisects the third side — and again it is the theorem with the ratio set to .
Result 4 — a strip between the parallel sides. In trapezium with , if lies on and on with , then
Proof idea: join , meeting at . Apply the theorem to with , then to with , and the two ratios both equal .
One construction idea covers all four results. Whenever a configuration has parallel lines but no obvious triangle, draw a diagonal or a line through the point of interest to create triangles the theorem can act on. That is the single technique this section is teaching, and it is worth practising deliberately — the algebra afterwards is always short.
Result 1 — the diagonals of a trapezium. In trapezium with , the diagonals meet at . Then
How to prove it. Draw a line through parallel to , meeting at . Since , this line is parallel to both. In , the line is parallel to , so by the theorem
and in , the line is parallel to , so
**Both equal the same ratio , so they equal each other.** That is the whole proof, and the construction line through is the only idea in it.
Worked example — an algebraic version. In trapezium with , the diagonals meet at . Given , , and , find .
Expanding both sides:
Now the rejection, and it is the point of the question. If then , a negative length, which is impossible. **So .
Check with **: , , , , and . Correct, and every length is positive.
Result 2 — mid-points. If and are the mid-points of and in , then
so by the converse of the theorem, .
That is the mid-point theorem, recovered as a special case. The Basic Proportionality Theorem is the general statement and the mid-point theorem is what it says when the ratio happens to be — which is a good example of a familiar result turning out to be a corollary of a stronger one.
Result 3 — a line through one mid-point. If is the mid-point of and the line through parallel to meets at , then
so ** is the mid-point of .** A line through the mid-point of one side, parallel to another side, bisects the third side — and again it is the theorem with the ratio set to .
Result 4 — a strip between the parallel sides. In trapezium with , if lies on and on with , then
Proof idea: join , meeting at . Apply the theorem to with , then to with , and the two ratios both equal .
One construction idea covers all four results. Whenever a configuration has parallel lines but no obvious triangle, draw a diagonal or a line through the point of interest to create triangles the theorem can act on. That is the single technique this section is teaching, and it is worth practising deliberately — the algebra afterwards is always short.
Exam tip
What does a full-mark similarity answer look like?
Name the theorem you are using, name the triangle you are using it in, and keep the correspondence of letters consistent. Marks here are for justification, not for arithmetic.
- **Write "by the Basic Proportionality Theorem in , since " — the theorem, the triangle and the parallel condition, all three
- Use the converse by name** when proving parallelism, and state the conclusion as "hence "
- Keep both fractions the same kind: part-to-part or part-to-whole, never one of each
- Choose the form that matches the given lengths. If and are given, use and skip a subtraction
- Compare fractions by cross-multiplication rather than by decimals, so nothing turns on rounding
- **In the letter order is a claim** — corresponds to , and every ratio must follow that order
- Check that every length in your answer is positive and reject any value of the unknown that makes one negative
- Give both conditions when asked what makes polygons similar, with a counterexample for each
- Draw and label the figure, marking equal ratios and the parallel sides
The misconception to name. Similar does not mean "almost congruent" or "roughly the same". It is an exact statement: equal corresponding angles together with a single scale factor shared by every pair of corresponding sides. Two rectangles of and have every angle equal and are not similar at all, so "looks alike" is not a test.
A second trap. Assuming the Basic Proportionality Theorem needs the parallel line to pass through mid-points, or to be drawn in some particular place. It holds for every parallel line cutting the other two sides, and the mid-point case is just the one where the ratio is . A student who only recognises the mid-point version will miss most of the questions in this chapter.
- **Write "by the Basic Proportionality Theorem in , since " — the theorem, the triangle and the parallel condition, all three
- Use the converse by name** when proving parallelism, and state the conclusion as "hence "
- Keep both fractions the same kind: part-to-part or part-to-whole, never one of each
- Choose the form that matches the given lengths. If and are given, use and skip a subtraction
- Compare fractions by cross-multiplication rather than by decimals, so nothing turns on rounding
- **In the letter order is a claim** — corresponds to , and every ratio must follow that order
- Check that every length in your answer is positive and reject any value of the unknown that makes one negative
- Give both conditions when asked what makes polygons similar, with a counterexample for each
- Draw and label the figure, marking equal ratios and the parallel sides
The misconception to name. Similar does not mean "almost congruent" or "roughly the same". It is an exact statement: equal corresponding angles together with a single scale factor shared by every pair of corresponding sides. Two rectangles of and have every angle equal and are not similar at all, so "looks alike" is not a test.
A second trap. Assuming the Basic Proportionality Theorem needs the parallel line to pass through mid-points, or to be drawn in some particular place. It holds for every parallel line cutting the other two sides, and the mid-point case is just the one where the ratio is . A student who only recognises the mid-point version will miss most of the questions in this chapter.
Did you know
Why can one theorem measure a tower you cannot climb?
The Basic Proportionality Theorem says something quietly powerful: a ratio of lengths survives a change of scale. The triangle can be tiny or enormous, and the two ratios stay equal.
Which means a ratio measured on something small can be transferred to something large. You never need to reach the far end of the thing you are measuring — you only need a ratio.
That is the principle behind every practical measurement in this chapter and in the next.
- A tower's height from its shadow. A metre stick and its own shadow give a ratio; the tower's shadow then gives the tower
- The width of a river. Pace out a small triangle on the bank whose shape matches a large one spanning the water
- A map or a scale drawing. Every length on the paper is one fixed multiple of the real length, which is exactly a similarity
- A photograph enlarged for printing. The same scale factor applies to every dimension, which is why an enlargement does not distort
And it explains something about photographs that is easy to miss. Enlarging a photograph to cannot be done without cutting or stretching, because while . Those rectangles are not similar, which is precisely the counterexample from the second section appearing in a print shop. A enlarges cleanly to and to nothing else.
The same reasoning explains why a triangle is the shape used for all of it. For a rectangle, knowing the angles tells you nothing about the sides. For a triangle, the angles alone fix the shape completely — which is why two triangles with equal angles are automatically similar, and why a surveyor can copy a large triangle as a small one just by copying its angles.
One last observation about what the theorem does not say. It gives you the ratio in which the two sides are divided; it says nothing about the length of the parallel line itself. That needs the similarity of the two triangles, which is the subject of Part 2 — and there the result finally lets you find . The theorem divides the sides; similarity measures the crossing line.
Which means a ratio measured on something small can be transferred to something large. You never need to reach the far end of the thing you are measuring — you only need a ratio.
That is the principle behind every practical measurement in this chapter and in the next.
- A tower's height from its shadow. A metre stick and its own shadow give a ratio; the tower's shadow then gives the tower
- The width of a river. Pace out a small triangle on the bank whose shape matches a large one spanning the water
- A map or a scale drawing. Every length on the paper is one fixed multiple of the real length, which is exactly a similarity
- A photograph enlarged for printing. The same scale factor applies to every dimension, which is why an enlargement does not distort
And it explains something about photographs that is easy to miss. Enlarging a photograph to cannot be done without cutting or stretching, because while . Those rectangles are not similar, which is precisely the counterexample from the second section appearing in a print shop. A enlarges cleanly to and to nothing else.
The same reasoning explains why a triangle is the shape used for all of it. For a rectangle, knowing the angles tells you nothing about the sides. For a triangle, the angles alone fix the shape completely — which is why two triangles with equal angles are automatically similar, and why a surveyor can copy a large triangle as a small one just by copying its angles.
One last observation about what the theorem does not say. It gives you the ratio in which the two sides are divided; it says nothing about the length of the parallel line itself. That needs the similarity of the two triangles, which is the subject of Part 2 — and there the result finally lets you find . The theorem divides the sides; similarity measures the crossing line.
Exam relevance
How does the proportionality theorem prepare you for JEE?
This is foundation work for Class 11 Straight Lines and Trigonometry, and for the coordinate-geometry and vector problems in JEE Main and JEE Advanced.
Where the division-in-a-ratio idea leads. Class 11 formalises it as the section formula in coordinate geometry and as the position-vector form in Class 12 vectors: a point dividing a segment in the ratio has coordinates built from that ratio. The theorem you prove here is the geometric fact that the section formula computes, and JEE questions on the centroid, on internal and external division, and on collinearity all read a ratio off a figure exactly this way.
Where the trapezium-diagonal result leads. It is the standard proof that the diagonals of a trapezium divide each other proportionally, and the same reasoning reappears in Vectors as the condition for two segments to intersect in given ratios. JEE Advanced sets it as a vector problem, where the construction line becomes a linear combination.
Where the area argument in the proof leads. "Triangles with the same height have areas in the ratio of their bases" becomes a tool for computing area ratios in coordinate geometry and, in Class 12, for the ratio in which a line divides a region under a curve. The area-ratio step is used far more often than the theorem it proves here.
Where the two similarity conditions lead. Class 11 Trigonometry rests on the fact that the ratio of two sides of a right triangle depends only on its angles — which is exactly the statement that equiangular triangles are similar. Every trigonometric ratio is well defined because of this chapter, and that connection is worth seeing explicitly: without similarity, would not be a single number.
Where the scale-factor idea leads. Class 11 Conic Sections and Class 12 transformations use scaling directly, and the result that all circles are similar is why the eccentricity of a circle does not depend on its radius.
Question types to expect. At this level: state the similarity conditions with counterexamples, prove the theorem, find a length, prove a line is parallel, and apply it in a trapezium. In competitive papers: ratio division on coordinates and vectors, concurrency and collinearity, and area-ratio problems.
The single trap that costs marks. Mixing a part-to-part ratio with a part-to-whole ratio. ** and are both true, but is not — and the same error at JEE level appears as confusing internal with external division in the section formula.
A second trap. Writing a similarity statement with the letters in the wrong order. fixes which vertex matches which**, and every ratio afterwards inherits that order. In coordinate and vector problems the equivalent slip is taking the ratio as instead of , which lands you at a different point entirely.
Board versus competitive emphasis. The CBSE paper marks the named theorem, the labelled figure, the stated parallel condition and the conclusion; a competitive paper marks the coordinates or the ratio. The transferable habit is writing down which ratio is equal to which before substituting any number — because in both settings the arithmetic is trivial and the correspondence is where everything is won or lost.
Where the division-in-a-ratio idea leads. Class 11 formalises it as the section formula in coordinate geometry and as the position-vector form in Class 12 vectors: a point dividing a segment in the ratio has coordinates built from that ratio. The theorem you prove here is the geometric fact that the section formula computes, and JEE questions on the centroid, on internal and external division, and on collinearity all read a ratio off a figure exactly this way.
Where the trapezium-diagonal result leads. It is the standard proof that the diagonals of a trapezium divide each other proportionally, and the same reasoning reappears in Vectors as the condition for two segments to intersect in given ratios. JEE Advanced sets it as a vector problem, where the construction line becomes a linear combination.
Where the area argument in the proof leads. "Triangles with the same height have areas in the ratio of their bases" becomes a tool for computing area ratios in coordinate geometry and, in Class 12, for the ratio in which a line divides a region under a curve. The area-ratio step is used far more often than the theorem it proves here.
Where the two similarity conditions lead. Class 11 Trigonometry rests on the fact that the ratio of two sides of a right triangle depends only on its angles — which is exactly the statement that equiangular triangles are similar. Every trigonometric ratio is well defined because of this chapter, and that connection is worth seeing explicitly: without similarity, would not be a single number.
Where the scale-factor idea leads. Class 11 Conic Sections and Class 12 transformations use scaling directly, and the result that all circles are similar is why the eccentricity of a circle does not depend on its radius.
Question types to expect. At this level: state the similarity conditions with counterexamples, prove the theorem, find a length, prove a line is parallel, and apply it in a trapezium. In competitive papers: ratio division on coordinates and vectors, concurrency and collinearity, and area-ratio problems.
The single trap that costs marks. Mixing a part-to-part ratio with a part-to-whole ratio. ** and are both true, but is not — and the same error at JEE level appears as confusing internal with external division in the section formula.
A second trap. Writing a similarity statement with the letters in the wrong order. fixes which vertex matches which**, and every ratio afterwards inherits that order. In coordinate and vector problems the equivalent slip is taking the ratio as instead of , which lands you at a different point entirely.
Board versus competitive emphasis. The CBSE paper marks the named theorem, the labelled figure, the stated parallel condition and the conclusion; a competitive paper marks the coordinates or the ratio. The transferable habit is writing down which ratio is equal to which before substituting any number — because in both settings the arithmetic is trivial and the correspondence is where everything is won or lost.
Key takeaways
What must you be able to do from this part?
One theorem, one converse and two conditions.
- Congruent means same shape and same size; similar means same shape, with size allowed to differ
- Every pair of congruent figures is similar, with scale factor
- Two polygons are similar only if their corresponding angles are equal and their corresponding sides are proportional
- Both conditions are needed: and rectangles have equal angles and are not similar; a square and a rhombus of the same side have proportional sides and are not similar
- All circles, all squares and all equilateral triangles are similar; two rectangles, rhombuses or isosceles triangles need not be
- For triangles only, either condition forces the other
- **In the letter order fixes the correspondence, and every ratio must follow it
- Basic Proportionality Theorem**: if in then
- The proof compares areas, and the parallel condition is used exactly once — to make the areas of and equal
- Equivalent forms: and — pick the one matching the given lengths
- The converse proves parallelism: equal ratios imply
- Never mix a part-to-part ratio with a part-to-whole ratio
- Compare fractions by cross-multiplication, so against becomes against
- Trapezium diagonals: with meeting at , — proved by drawing a line through parallel to
- Reject any value making a length negative, as is rejected in the trapezium example leaving
- **The mid-point theorem is this theorem with the ratio equal to **
The sharpest self-test is a figure you draw yourself. Sketch any triangle, draw a parallel line across it wherever you like, measure the four pieces with a ruler, and see how close your two ratios come — then explain the difference in terms of measurement error rather than doubting the theorem.
- Congruent means same shape and same size; similar means same shape, with size allowed to differ
- Every pair of congruent figures is similar, with scale factor
- Two polygons are similar only if their corresponding angles are equal and their corresponding sides are proportional
- Both conditions are needed: and rectangles have equal angles and are not similar; a square and a rhombus of the same side have proportional sides and are not similar
- All circles, all squares and all equilateral triangles are similar; two rectangles, rhombuses or isosceles triangles need not be
- For triangles only, either condition forces the other
- **In the letter order fixes the correspondence, and every ratio must follow it
- Basic Proportionality Theorem**: if in then
- The proof compares areas, and the parallel condition is used exactly once — to make the areas of and equal
- Equivalent forms: and — pick the one matching the given lengths
- The converse proves parallelism: equal ratios imply
- Never mix a part-to-part ratio with a part-to-whole ratio
- Compare fractions by cross-multiplication, so against becomes against
- Trapezium diagonals: with meeting at , — proved by drawing a line through parallel to
- Reject any value making a length negative, as is rejected in the trapezium example leaving
- **The mid-point theorem is this theorem with the ratio equal to **
The sharpest self-test is a figure you draw yourself. Sketch any triangle, draw a parallel line across it wherever you like, measure the four pieces with a ruler, and see how close your two ratios come — then explain the difference in terms of measurement error rather than doubting the theorem.