Question

In the adjoining figure, ABCD is a parallelogram and X, Y are the points on the diagonal BD such that DX = BY
Prove that
(i) CXAY is a parallelogram,
(ii) ∆ADX ≅ ∆CBY and ∆ABY ≅ ∆CDX,and
(iii) AX = CY and CX = AY.
​

Answer 1

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

In the adjoining figure, ABCD is a parallelogram and X, Y are the points on the diagonal BD such that DX = BY

Prove that

(i) CXAY is a parallelogram,

(ii) ∆ADX ≅ ∆CBY and ∆ABY ≅ ∆CDX,and

(iii) AX = CY and CX = AY.

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

━━━━━━━━━━━━━━━

(i) Join AC, meeting BD at O.

Since the diagonal of s parallelogram bisect each other,

we have OA = OC and OD = OB.

Now,OD = OB and DX = BY

⇒OD - DX = OB - BY ⇒ OX = OY.

Now, OA = OC and OX = OY.

CXAY is a quadrilateral whose diagonals bisect each other.

∴CXAY is a ||gm.

━━━━━━━━━━━━━━━

(ii) and (iii) In ∆ADX and CBY, we have

AD = BC (opp. sides of a ||gm)

∠ADX = ∠CBY

DX = BY (given)

∴∆ADX ≅ ∆CBY (SAS-criterion)

And so, AX = CY (c.p.c.t.)

Similarly, ∆ABY ≅ ∆CDX and so CX = AY (c.p.c.t.)

━━━━━━━━━━━━━━━