$ABCD$ 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.

Given:

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

To do:

We have to prove that $EFGH$ is a square.

Solution:

Let $AE = BF = CG = DH = x$ and $BE = CF = DG = AH = y$

In $\triangle AEH$ and $\triangle BFE$,

$AE = BF$           (Given)

$\angle A = \angle B$

$AH = BE$

Therefore, by SAS axiom,

$\triangle AEH \cong \triangle BFE$

This implies,

$\angle 1 = \angle 2$

$\angle 3 = \angle 4$

$\angle 1 + \angle 3 = 90^o$

$\angle 2 + \angle 4 = 90^o$

$\angle 1 + \angle 2 + \angle 3 + \angle 4 = 90^o + 90^o = 180^o$

$\angle 1 + \angle 4 + \angle 1 + \angle 4 = 180^o$

$2(\angle 1 + \angle 4) = 180^o$

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

Therefore,

$\angle HEF = 180^o - 90^o = 90^o$

Similarly,

$\angle F = \angle G = \angle H = 90^o$

Here, the sides of quadrilateral $EFGH$ are equal and each angle is equal to $90^o$

This implies,

$EFGH$ is a square.

Hence proved.

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Updated on: 10-Oct-2022

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