Mathematics
In the given figure, AY ⊥ ZY nd BY ⊥ XY such that AY = ZY and BY = XY. Prove that AB = ZX.
Answer
⇒ ∠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.
Related Questions
In the given figure, ∠BAC = ∠CDB and ∠BCA = ∠CBD. Prove that AB = CD.
In the given figure, ∠ABD = ∠EBC, BD = BC and ∠ACB = ∠EDB. Prove that AB = BE.
In the given figure, ABCD is a square and △PAB is an equilateral triangle.
(i) Prove that △APD ≅ △BPC.
(ii) Show that ∠DPC = 15°.
In the given figure, in △ABC, ∠B = 90°. If ABPQ and ACRS are squares, prove that:
(i) △ACQ ≅ △ABS
(ii) CQ = BS.
