ABCD is a square. E, F, G and H are the mid points of 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 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

Hence quadrilateral is square