In the given figure, AY ⊥ ZY nd BY ⊥ XY such that AY = ZY and BY = XY. Prove that AB = ZX.

⇒ ∠AYZ = ∠XYB [Each equal to 90°]

Adding ∠AYX on both L.H.S and R.H.S, we have:

⇒ ∠AYZ + ∠AYX = ∠XYB + ∠AYX

⇒ ∠XYZ = ∠AYB

In △XYZ and △AYB,

⇒ ZY = AY [Given]

⇒ XY = BY [From figure]

⇒ ∠XYZ = ∠AYB [Proved above]

∴ △XYZ ≅ △AYB (By S.A.S axiom)

⇒ AB = ZX [Corresponding parts of congruent triangles are equal]

Hence, proved that AB = ZX.