Three Measurements Are Enough to Prove Two Triangles Match

No — three of the six are enough, provided they are the right three. A triangle has three sides and three angles, but certain trios lock the triangle completely, so matching only those proves the two are identical copies.

This page covers everything in the CBSE Class 7 Mathematics chapter on congruence: what congruence means, writing a congruence statement, the SSS, SAS and ASA criteria, the RHS criterion for right-angled triangles, and using congruence to prove the angle properties of isosceles and equilateral triangles.

Two figures are congruent if one can be placed exactly over the other, covering it completely with nothing left over.

The practical test is tracing. Copy one figure onto tracing paper, then slide, turn or flip it onto the other. If it fits exactly, the two are congruent, and this is written with the symbol $\cong$.

- Two line segments are congruent when they have the same length.
- Two angles are congruent when they have the same measure, however long their arms happen to be drawn.

Two identical set-squares from a geometry box stay congruent even when one is turned over, because sliding, turning and flipping change no length and no angle.

The order of the letters in a congruence statement carries real information. Writing $△ABC\cong △DEF$ says that A matches D, B matches E and C matches F — so AB = DE, BC = EF, CA = FD, and $\angle A=\angle D$.

That is why $△ABC\cong △DEF$ and $△ABC\cong △EFD$ say different things, and only one of them can be true for a given pair of triangles.

Each criterion names three measurements that fix a triangle completely, so matching them is enough.

SSS — three pairs of equal sides. If AB = DE, BC = EF and CA = FD, then $△ABC\cong △DEF$.

SAS — two pairs of equal sides together with the included angle, meaning the angle lying between those two sides.

ASA — two pairs of equal angles together with the included side, the side lying between those two angles.

Worked example. In $△ABC$: AB = 5 cm, BC = 7 cm and $\angle B=50°$. In $△PQR$: PQ = 5 cm, QR = 7 cm and $\angle Q=50°$. In both triangles the $50°$ angle sits between the 5 cm and 7 cm sides, so the triangles are congruent by SAS, and you can then conclude CA = RP without measuring.

The word included is what makes or breaks these criteria. Two sides and a non-included angle can fit two genuinely different triangles, which is why that combination is not on the list.

The link back to construction is worth seeing: these are the same trios that let you build exactly one triangle with ruler and compasses, and for the same reason.

RHS applies only to right-angled triangles: if the hypotenuse and one other side match, the triangles are congruent.

Worked example. Two right-angled triangles each have hypotenuse 13 cm and one leg 5 cm. They are congruent by RHS, and the third side is forced in both, since

$$13^2-5^2=169-25=144=12^2$$

giving 12 cm each time. The right angle removes any choice about where the third vertex can sit.

Three equal angles, by contrast, prove nothing about size. An equilateral triangle of side 2 cm and one of side 20 cm have exactly the same three angles of $60°$, yet one fits on your thumbnail and the other does not fit on the page.

So equal angles give the same shape but not the same size. Congruence needs at least one pair of equal sides somewhere in the trio, because only a side can set the scale.

A square floor tile and a much larger square tile make the same point: identical angles of $90°$, completely different tiles.

By splitting the triangle into two halves and showing the halves are congruent.

Take $△ABC$ with AB = AC. Draw AD, the bisector of $\angle A$, meeting BC at D. Now compare $△ABD$ and $△ACD$:

- AB = AC (given)
- $\angle BAD=\angle CAD$ (AD bisects the angle)
- AD = AD (the same side in both)

That is SAS, so $△ABD\cong △ACD$, and therefore $\angle B=\angle C$. The base angles are equal because the two halves are identical copies.

Applied: if the unequal angle of an isosceles triangle measures $40°$, the two base angles share the remaining $180-40=140°$, giving $70°$ each.

For an equilateral triangle all three sides are equal, so the same argument makes all three angles equal, and since they total $180°$ each must be $\frac{180}{3}=60°$.

The statement also works backwards, which questions often need: if two angles of a triangle are equal, then the sides opposite them are equal, so the triangle is isosceles.