Congruent Triangles

Compare the three pairs of sides: both triangles have sides of $3\text{ cm}$, $5\text{ cm}$ and $7\text{ cm}$.

All three pairs of corresponding sides are equal, so the triangles are congruent by the Side-Side-Side (SSS) rule.

The green triangle has been turned round, but rotating or reflecting a shape does not change any of its side lengths, so it is still congruent to the purple one.

Both triangles have a $4\text{cm}$ side and a $5\text{cm}$ side with the $70^{\circ }$ angle between them, so they match by the Side-Angle-Side (SAS) rule.

The pink triangle is drawn turned round compared with the blue one, but rotating a triangle does not change any of its side lengths or angles, so the two triangles are still congruent.

Both triangles have a right angle, the same $10\text{cm}$ hypotenuse and the same $6\text{cm}$ side, so the Right angle-Hypotenuse-Side (RHS) rule applies and they are congruent.

You do not need all six measurements. Pythagoras fixes the third side at $\sqrt{10^2-6^2}=\sqrt{64}=8$ cm, so these measurements can only make one triangle.

Both triangles have angles of $80^{\circ }$ and $50^{\circ }$ with the $5\text{cm}$ side between those two angles, so they match by the Angle-Side-Angle (ASA) rule.

The green triangle is drawn turned round compared with the orange one, but rotating a triangle leaves every side and angle unchanged, so the two triangles are congruent.

AAA is not enough to prove congruence, because three angles fix only the shape of a triangle, not its size. Two triangles can have exactly the same three angles but different side lengths - they are similar, not necessarily congruent.

ASA, SAS and SSS each include at least one side, which fixes the size as well as the shape. SSS still proves congruence when one triangle is a rotation or mirror image of the other, and in SAS the angle used must be the one between the two given sides.

The $40^{\circ }$ angle is opposite the $5\text{cm}$ side, not between the two given sides, so this is SSA.

By the sine rule, $\sin (B)=\frac{7\times \sin (40^{\circ })}{5}\approx 0.900$, which gives $B\approx 64.1^{\circ }$ or $B\approx 115.9^{\circ }$. Both work, because $40^{\circ }+115.9^{\circ }$ is still less than $180^{\circ }$, so two genuinely different triangles fit the same three measurements. SSA does not guarantee congruence.