Question

Prove that the diagonals of a square divide the square into four congurent triangles.

Answer 1

AB$=$BC$=$CD$=$DA

Angle B$=$C$=$D$=$A$={90}^o$

Let the centre be $0$

Then in each triangle one side is equal we know that the diagonals perpendicularly based in square.

So angle $0$ is equal in way triangle and one angle is equal in every triangle other than $0$ so by ASA

All $4$ triangles are congruent.