ABCD is a parallelogram and line-segments AX, CY bisect the angles A and C respectively. Prove that, AX || CY.
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Given: A ||gm ABCD and line segments AX, CY bisect the angles A and C bisect ∠A and ∠C respectively.
To prove: AX || CY
Proof:
∠A = ∠C (opposite ∠s of a ||gm)
∴ 1/2 ∠A = 1/2∠C
i.e. ∠XAY = ∠XCY (AX and CY are bisectors of ∠s A and C respectively)
Now,
AB || DC and CY intersect them
∴ ∠XAY + ∠AXC = 180° (Co-interrior angles) ...(i)
AB || DC and CY intersects them
∴ ∠XCY + ∠AYC = 180° (Co-interrior angles).....(ii)
From equations (i) and (ii) we get that,
∠XAY + ∠AXC = ∠XCY + ∠AYC
⇒ ∠XAY + ∠AXC = ∠XAY + ∠AYC (∵ ∠XYC = ∠XAY, PROVED)
∴ ∠AXC = ∠AYC
Such that, opposite angles of quadrilateral AYCX are equal.
Hence, AYCX is ||gm ⇒ AX || CY
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