Question

$ABCD$ is a square. $E,F,G$ and $H$ are the mid point $AB,BC,CD$ and $DA$ respectively. Such that $AE=BF=CG=DH$. Prove that $EFGH$ is a square

Answer 1

Given : $ABCD$ is a square. $E,F,G$ and $H$ are the midpoints of $AB,BC,CD$ and $DA$ respectively. Such that $AE=BF=CG=DH$

To show: $EFGH$ is a square

Proof :

$AE=BF=CG=DH$

Therefore, $BE=CF=DG=AH$

In trinagle $AEH$ and triangle $BFE$

$AE=BF$ (Given)

$\angle A=\angle B$ (each equal to ${90}^{\circ })$

$AH=BE$

By SAS criterion of congruency, triangle $AEH$ is congruent to triangle $BFE.$

$\Rightarrow EH=BF$ (By CPCT)

Similarity, $EH=HG=GF=FE$

Now, $\angle AEH=\angle BFE$ and $\angle AHE=\angle BEF$

But, $\angle AEH+\angle AHE={90}^{\circ }$ and $\angle BFE+\angle BFE={90}^{\circ }$

$\Rightarrow \angle AEH+\angle AHE+\angle BFE+\angle BFE={90}^{\circ }+{90}^{\circ }$

$\Rightarrow \angle AEH+\angle BEF+\angle AEH+\angle BEF={90}^{\circ }+{90}^{\circ }$

$\Rightarrow 2(\angle AEH+\angle BEF)={180}^{\circ }$

$\Rightarrow \angle AEH+\angle BEF={90}^{\circ }$

$\Rightarrow \angle HEF={90}^{\circ }$

Similarly, $\angle EFG=\angle FGH=\angle GHE={90}^{\circ }$

Therefore, $EFGH$ is a square