Question

Prove that the diagonals of square divide it into four congruent triangles.

Answer 1

To prove: The diagonals of square divide it into four congruent triangles.

Proof:

Let $ABCD$ is a square.

Diagonals $AC$ and $BD$ intersects at the point $O$.

Diagonals divides the square into four triangle $\Delta AOB,\Delta BOC,\Delta COD$ and $\Delta DOA$.

Consider the triangles $\Delta AOB$ and $\Delta BOC$
$AO=CO$ (Diagonals of a square bisect each other.)
$OB=OB$ (Common)
$AB=CB$ (All sides of a square are equal.)
By SSS congruency criteria,

$\Delta AOB\cong \Delta BOC$ ---(1)

Similarly, $\Delta BOC\cong \Delta COD$

$\Delta COD\cong \Delta DOA$

$\Delta DOA\cong \Delta AOB$

Therefore, $\Delta AOB\cong \Delta BOC\cong \Delta COD\cong \Delta DOA$
Thus, we can say that the diagonals of a square divide it into four congruent triangles.

Hence, proved.