Question

ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.

Answer 1

Given : In square ABCD

E, F, G and H are the points on AB, BC, CD and DA respectively suct that AE = BF= CG = DH

To prove : EFGH is a square

Proof : E,F, G and H are points on the sides AB,BC, CA and DA respectively such that

AE = BF =CG=DH =x (suppose)

Then BE =CF=DG=AH=y (suppose)

Now in $\Delta AEHand\Delta BFE,$

AE = BF (given)

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

AH = BE (proved)

$\therefore \Delta AEH\cong \Delta BFE$ (SAS criterion)

$\therefore \angle 1=\angle 2and\angle 3=\angle 4$ (c.p.c.t.)

But $\angle 1+\angle 3={90}^{\circ }and\angle 2+\angle 4={90}^{\circ }(\angle A=\angle B={90}^{\circ })$

$\Rightarrow \angle 1+\angle 2+\angle 3+\angle 4={90}^{\circ }+{90}^{\circ }={180}^{\circ }$

$\Rightarrow \angle 1+\angle 4+\angle 1+\angle 4={180}^{\circ }$

$\Rightarrow 2(\angle 1+\angle 4)={180}^{\circ }$

$\Rightarrow \angle 1+\angle 4=\frac{{180}^{\circ }}{2}={90}^{\circ }$

$\therefore \angle HEF={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

Similarly, we can prove that

$\angle F=\angle G=\angle H={90}^{\circ }$

Since sides of quad. EFGH is are equal and each angle is of ${90}^{\circ }$

$\therefore$ EFGH is a square.