Two Matching Angles Are Enough to Fix the Shape of a Triangle Completely

Try to draw a triangle with angles of $50^{\circ }$, $60^{\circ }$ and $70^{\circ }$ that is a different shape from someone else's triangle with the same three angles. You cannot. You can draw a bigger one or a smaller one, but the shape is forced.

Now try the same with a rectangle. Every rectangle has four right angles, and yet a $2$ cm by $3$ cm rectangle and a $2$ cm by $10$ cm rectangle look nothing alike. Angles fix the shape of a triangle and of no other polygon.

That is why the previous part ended with two conditions for similarity — equal angles and proportional sides — while this part can offer three shortcuts that need only half the information:

- AA — two pairs of equal angles are enough
- SSS — three pairs of proportional sides are enough
- SAS — two pairs of proportional sides with the included angles equal are enough

Each of these gives you the full similarity, and from there every remaining ratio follows for free. That is what makes the criteria worth having: you check two or three things and then use six.

And the ratio extends further than the sides themselves. If two triangles are similar with scale factor $k$, then their altitudes, medians, angle bisectors and perimeters are all in the same ratio $k$. One number describes the whole enlargement, which is why a single measurement on a small triangle can give a length on a large one.

That last point is where the chapter meets the real world. A metre stick and its shadow give you a ratio, and a tower's shadow then gives you the tower — without anyone climbing anything.

This page covers the second part of the CBSE Class 10 Maths chapter on triangles: the AA, SSS and SAS similarity criteria, proportionality of corresponding lines, and applications to heights and distances.

The AAA criterion says that if the three angles of one triangle equal the three angles of another, the triangles are similar. The AA criterion notes that two pairs suffice, because the third pair is then automatic.

Why two pairs are enough. The angles of any triangle add to $180^{\circ }$. So if $\angle A=\angle P$ and $\angle B=\angle Q$, then

$$\angle C=180^{\circ }-\angle A-\angle B=180^{\circ }-\angle P-\angle Q=\angle R$$

The third equality is forced, which is why AA and AAA are the same criterion in practice and why you should quote AA — checking a third angle earns nothing.

Worked example 1 — the standard configuration. In $△ABC$, the points $D$ on $AB$ and $E$ on $AC$ satisfy $DE\parallel BC$. Given $AD=2$ cm, $AB=6$ cm and $BC=9$ cm, find $DE$.

First establish the similarity. Since $DE\parallel BC$ with $AB$ as a transversal, $\angle ADE=\angle ABC$ as corresponding angles. And $\angle A$ is common to both triangles. **So by the AA criterion, $△ADE\sim △ABC$.

Then use the proportionality of corresponding sides:**

$$\frac{AD}{AB}=\frac{DE}{BC}⟹\frac{2}{6}=\frac{DE}{9}$$
$$DE=\frac{9\times 2}{6}=3\text{ cm}$$

Check the scale factor. It is $\frac{2}{6}=\frac{1}{3}$, so every length in the small triangle should be a third of its partner — and $3$ is indeed a third of $9$. Correct.

Notice what this adds to Part 1. The Basic Proportionality Theorem told you how $DE$ divides the two sides; **it said nothing about the length of $DE$ itself. Similarity supplies that, and this is the gap the criteria fill.

Worked example 2 — the perpendicular to the hypotenuse.** In $△ACB$, right-angled at $C$, a perpendicular $CD$ is drawn from $C$ to the hypotenuse $AB$. Show that $C{D}^2=AD\times DB$, and find $CD$ when $AD=4$ cm and $DB=9$ cm.

Establish the similarity. In triangles $ACD$ and $CBD$:

- $\angle ADC=\angle CDB=90^{\circ }$
- $\angle DAC=\angle DCB$, because both are the complement of $\angle ACD$ in their own right triangle

**So $△ACD\sim △CBD$ by the AA criterion.** Matching the corresponding sides in that order,

$$\frac{AD}{CD}=\frac{CD}{DB}⟹C{D}^2=AD\times DB$$

With the given lengths,

$$C{D}^2=4\times 9=36⟹CD=6\text{ cm}$$

Check by Pythagoras. Here $AB=4+9=13$ cm. From the similar triangles, $A{C}^2=AD\times AB=4\times 13=52$ and $B{C}^2=DB\times AB=9\times 13=117$. Then

$$A{C}^2+B{C}^2=52+117=169=13^2=A{B}^2$$

The Pythagoras theorem is satisfied exactly, which confirms the whole configuration.

A second check using area. The area of $△ACB$ is $\frac{1}{2}\times AB\times CD=\frac{1}{2}\times 13\times 6=39$ cm${}^2$. Using the legs instead, it is $\frac{1}{2}\times \sqrt{52}\times \sqrt{117}=\frac{1}{2}\sqrt{6084}=\frac{78}{2}=39$ cm${}^2$. The same area by two routes.

Why the AA criterion is the one you will use most. Angles appear free of charge in any figure with parallel lines, common vertices, right angles, or angles in the same segment of a circle. You rarely have to measure anything to find two equal angles, whereas SSS and SAS need lengths. When a similarity question looks hard, look for two equal angles first.

When the question gives you sides rather than angles. SSS needs all three ratios equal; SAS needs two ratios equal and the angles between those sides equal.

The SSS criterion. If the three sides of one triangle are proportional to the three sides of another, the triangles are similar:

$$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}⟹△ABC\sim △PQR$$

Worked example 1. Are the triangles with sides $4$ cm, $6$ cm, $8$ cm and $6$ cm, $9$ cm, $12$ cm similar?

Pair the sides in increasing order — smallest with smallest, largest with largest — and compute the three ratios:

$$\frac{4}{6}=\frac{2}{3},\frac{6}{9}=\frac{2}{3},\frac{8}{12}=\frac{2}{3}$$

All three ratios are equal, so the triangles are similar by SSS, with scale factor $\frac{2}{3}$.

Worked example 2 — where it fails. Are the triangles with sides $3$ cm, $4$ cm, $5$ cm and $6$ cm, $8$ cm, $12$ cm similar?

$$\frac{3}{6}=\frac{1}{2},\frac{4}{8}=\frac{1}{2},\frac{5}{12}$$

The first two ratios agree and the third does not, so the triangles are not similar. Two out of three is not enough, and a student who stops after checking two ratios will get this wrong.

The SAS criterion. If two sides of one triangle are proportional to two sides of another and the angles included between those sides are equal, the triangles are similar.

Worked example 3. In triangles $ABC$ and $DEF$, $\angle A=\angle D=70^{\circ }$, $AB=4$ cm, $AC=6$ cm, $DE=6$ cm and $DF=9$ cm. Are they similar?

$$\frac{AB}{DE}=\frac{4}{6}=\frac{2}{3},\frac{AC}{DF}=\frac{6}{9}=\frac{2}{3}$$

**The two ratios are equal, and $\angle A$ and $\angle D$ are the angles between those very sides.** So by the SAS criterion, $△ABC\sim △DEF$.

The word "included" is doing the work here. $\angle A$ lies between $AB$ and $AC$, and $\angle D$ lies between $DE$ and $DF$. **Had the equal angles been $\angle B$ and $\angle E$ instead, the criterion would not apply, because those angles are not between the pairs of sides whose ratios were computed. An equal angle that is not the included one proves nothing, and that is the trap every SAS question is built around.

Worked example 4 — finding a side once similarity is established.** In the triangles of worked example 3, $BC=5$ cm. Find $EF$.

With $△ABC\sim △DEF$ and scale factor $\frac{2}{3}$,

$$\frac{BC}{EF}=\frac{2}{3}⟹EF=\frac{3\times 5}{2}=7.5\text{ cm}$$

That is the pay-off of a similarity criterion. You verified two ratios and one angle, and you are now entitled to the third ratio without measuring it.

How to choose the criterion. Read what the question gives you:

- Angles, parallel lines, right angles or a common angle — use AA
- All three sides of both triangles — use SSS
- Two sides of each and the angle between them — use SAS

And a warning about the congruence criteria. ASA, RHS and the plain SSS of congruence are not similarity criteria, and "SSA" is not a criterion at all in either setting. There is no AAA congruence and no SSA similarity, and quoting one of those is marked wrong even when the conclusion happens to be true.

Find or create two similar triangles, one of which you can measure completely. The ratio from the small triangle then gives the length you cannot reach.

Worked example 1 — a tower from shadows. A vertical pole $6$ m long casts a shadow $4$ m long on the ground, and at the same time a tower casts a shadow $28$ m long. Find the height of the tower.

Why the triangles are similar. The sun's rays arrive along parallel lines, so the angle of elevation of the sun is the same at both places. Each of the pole and the tower stands vertically, making a right angle with the ground. Two pairs of equal angles, so the two triangles are similar by AA.

Let the tower's height be $h$ metres. Matching height with height and shadow with shadow,

$$\frac{6}{4}=\frac{h}{28}⟹h=\frac{6\times 28}{4}=42\text{ m}$$

Check the ratios: $\frac{6}{4}=1.5$ and $\frac{42}{28}=1.5$. Correct.

The phrase "at the same time" is doing essential work. It guarantees the sun is at the same elevation for both measurements. Shadows measured at different times of day give different ratios, and a question that omits the phrase is a faulty question.

Worked example 2 — a shadow that is moving. A girl of height $90$ cm walks away from the base of a lamp-post at $1.2$ m/s. If the lamp is $3.6$ m above the ground, find the length of her shadow after $4$ seconds.

First convert her height: $90$ cm $=0.9$ m. In $4$ seconds she has walked

$$1.2\times 4=4.8\text{ m}$$

Let her shadow be $s$ metres long, measured from her feet. The lamp, the top of her head and the tip of her shadow lie on one straight line, so the small triangle formed by the girl and her shadow is similar to the large triangle formed by the lamp-post and the whole distance from its base to the shadow's tip. Both are right triangles sharing the angle at the shadow's tip, so AA applies. Hence

$$\frac{3.6}{0.9}=\frac{4.8+s}{s}$$
$$4s=4.8+s⟹3s=4.8⟹s=1.6\text{ m}$$

Check: $\frac{4.8+1.6}{1.6}=\frac{6.4}{1.6}=4$, and $\frac{3.6}{0.9}=4$. Correct.

The step that decides this problem is what goes in the denominator. The large triangle's base is the distance from the lamp-post to the shadow's tip, which is $4.8+s$, not $4.8$. Drawing the figure before writing the ratio is the only reliable guard, and it is where most attempts at this question go wrong.

Worked example 3 — a distance you cannot walk. To find the distance $AB$ across a pond, a surveyor picks a point $O$ on the bank, extends $AO$ beyond $O$ to $C$ so that $OC=\frac{1}{2}AO$, and extends $BO$ beyond $O$ to $D$ so that $OD=\frac{1}{2}BO$. She then measures $CD$ and finds it to be $26$ m. Find $AB$.

Why the triangles are similar. In triangles $OAB$ and $OCD$:

- $\frac{OA}{OC}=2$ and $\frac{OB}{OD}=2$, so two pairs of sides are proportional
- $\angle AOB=\angle COD$, because they are vertically opposite angles — and these are the included angles

**So $△OAB\sim △OCD$ by SAS.** Therefore

$$\frac{AB}{CD}=2⟹AB=2\times 26=52\text{ m}$$

The pond was never crossed. Every length actually measured — $OC$, $OD$ and $CD$ — was on dry land, and the vertically-opposite angle supplied the one thing that could not be measured.

Worked example 4 — a streetlight and two poles. Two vertical poles of heights $6$ m and $11$ m stand on level ground with their feet $12$ m apart. Find the distance between their tops.

This one needs Pythagoras rather than similarity, and it is worth seeing why. The horizontal separation is $12$ m and the vertical difference is $11-6=5$ m, so the distance between the tops is

$$\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ m}$$

Not every height-and-distance question is a similarity question. Similarity is the tool when two triangles of the same shape exist; Pythagoras is the tool when one right triangle has two known sides. Deciding which applies is part of the question, and the give-away is whether the problem hands you a ratio or a pair of perpendicular lengths.