Question

ABCD is a square. Mid point of AB is X and mid point of BC is Y. Triangles ADX and BAY are congruent. Angle DXA = angle AYB
Prove that : DX is perpendicular to AY.

I'll mark the answer as brainliest

Answer 1

In Δs DAX and ABY,
DA=AB
AX=BY
∠DAX=∠ABY
ΔDAX≅ΔABY
Hence ∠AXD=∠BYA
In quad XOYB,
∠XOY+∠OYB+∠YBX+∠BXO = 360
∠XOY+∠OYB+90+(180 - AXO) = 360
But ∠OYB=∠AXO  (as ∠AXD=∠BYA)
⇒∠XOY=90
Thus DX⊥AY
Cheers and don't forget to mark it as brainliest!!!

Answer 2

In Δs DAX and ABY,

DA=AB

AX=BY

∠DAX=∠ABY

ΔDAX≅ΔABY

Hence ∠AXD=∠BYA

In quad XOYB,

∠XOY+∠OYB+∠YBX+∠BXO = 360

∠XOY+∠OYB+90+(180 - AXO) = 360

But ∠OYB=∠AXO (as ∠AXD=∠BYA)

⇒∠XOY=90

Thus DX⊥AY