# ABCD is a square.

Given:- *ABCD is a square

*E, F, G and H are the mid points of AB, BC, CD and DA

respectively

*AE = BF = CG = DH

Formula used:- *Isosceles Δ property

If 2 sides of triangle are equal then

Corresponding angle will also be equal

*Properties of quadrilateral to be square

All sides are equal

All angles are 90°

Solution:-

In Δ AHE, Δ EBF, Δ FCG, Δ DHG

AE = BF = CG = DH [Given]

If;

AE=EB [E is midpoint of AB]

BF=FC [F is midpoint of BC]

CG=GD [G is midpoint of CD]

DH=HA [H is midpoint of DA]

AE = BF = CG = DH

On replacing every part we get;

EB=FC=GD=HA;

A+ B+ C+ D=90° [All angles of square are 90°]

Hence;

All triangles Δ AHE, Δ EBF, Δ FCG, Δ DHG

are congruent by SAS property

Δ AHE Δ EBF Δ FCG Δ DHG

HE=EF=FG=GH [All triangles are congruent]

In Δ AHE, Δ EBF, Δ FCG, Δ DHG

all sides of square are equal and after the midpoint of each sides

Every half side of square are equal to half of other sides.

HA=AE , EB=FB ,FC=GC ,HD=DG

All Δ AHE, Δ EBF, Δ FCG, Δ DHG are isosceles

as central angle of all triangle is 90°

It makes all Δ AHE, Δ EBF, Δ FCG, Δ DHG are right angle isosceles Δ

all corresponding angles of equal side will be 45°

AHE= BEF= CFG= DHG= AEH= BFE= CGF= DGH=45°

as AB is straight line

Then; AEH+ HEF+ BEF=180°

45°+ HEF+45° =180°

HEF=180° -90° =90°

Similarly ;

EFG=90°

FGH=90°

GHE=90°

Conclusion:-

All angles are 90° and all sides are equal of quadrilateral

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