Question

BY bisects angle b of triangle ABC. Prove that BZYX is a rhombus​

Answer 1

BZYX is a rhombus​  if BY bisects angle b of triangle AB & ZY ║ BC & BZ ║XY

Step-by-step explanation:

BX ║ ZY

=> ∠BYZ = ∠YBX  

BZ ║XY

& ∠BYX = ∠YBZ

BY bisects angle b of triangle ABC

=> ∠YBZ = ∠YBX  

=> ∠BYZ = ∠YBX   =  ∠BYX = ∠YBZ

now in Δ BZY

∠YBZ =  ∠BYZ

=> BZ = ZY

& in Δ BXY

∠YBX   =  ∠BYX

=> BX = XY

now Δ BZY & Δ BXY

BY = BY common

∠YBZ = ∠YBX

∠BYZ =  ∠BYX

=> Δ BZY ≅ Δ BXY

BZ = BX

YZ = YX

=> BX = XY = YZ = BZ

Hence BZYX is a rhombus​

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Answer 2

Step-by-step explanation:

this is the answer of the question