Question

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

Answer 1

Take the side of square as a 
Then, AB= a; BF = a/2. 
Hence, in triangle ABF, 
AF = sqrt (a^2 + (a/2)^2) (By Pythogaras theorem) 
= sqrt (5 a^2 /4) = a /2 x sqrt (5) 
Again sin A = ((a/2)/(a /2 x sqrt (5))) = 1/sqrt (5) 
cos A = ((a)/(a /2 x sqrt (5))) = 2 / sqrt (5) 

As you can see PQRS forms a square. So, angle P = 90° 
Now, in triangle APE, 
sin A = PE/AE---> PE = AE sin A = (a/2)(1/sqrt (5)) = a/(2 sqrt (5)) 
cos A = AP/AE---> AP = AE cos A = (a/2)(2/sqrt (5)) = a/( sqrt (5)) 

AF = AP+ PQ+QF 
a sqrt (5) / 2= a/(2 sqrt (5)) + PQ+ a/( sqrt (5)) 
PQ = a sqrt (5) / 2 - ( a/(2 sqrt (5)) + a/( sqrt (5))) 
= (5a - (a+2a)/(2 sqrt (5))) = a/ sqrt (5) 

(Check any line and you can see this.. QF = AE)
Hence, side PQ = a/ sqrt (5) 
Area of PQRS = a^2 /5 
Area of ABCD = a^2 

The ratio of the areasPQRS : ABCD is (a^2 /5)/a^2 = 1/5 
= 0.2

Answer 2

Given AE=BF=CG=DH

⟹ So, EB=FC=GD=HA

In △s AEH and BFE,

AE=BF, AH=EB,

∠A=∠B (each ∠ = 90⁰)

∴ △AEH ≅ △BFE

⟹ EH=EF and ∠4= ∠2.

But ∠1 + ∠4 = 90⁰ ⟹ ∠1 + ∠2 = 90⁰

⟹ ∠HEF = 90⁰

And if ∠HEF = 90⁰ so, ∠EFG = 90⁰, ∠FGH = 90⁰ and ∠GHE = 90⁰.

Hence Proved.

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