Mathematics
ABCD is a parallelogram in which ∠A and ∠C are obtuse. Points X and Y are taken on diagonal BD such that ∠AXD = ∠CYB = 90°. Prove that : XA = YC.
Triangles
2 Likes
Answer
AD || BC and BD is the transversal.
⇒ ∠ADX = ∠CBY (Alternate interior angles are equal)
In △XAD and △YCB,
⇒ ∠ADX = ∠CBY (Proved above)
⇒ ∠AXD = ∠BYC (Both equal to 90°)
⇒ AD = BC (Opposite sides of a parallelogram are equal)
∴ △XAD ≅ △YCB (By A.A.S. axiom)
⇒ XA = YC (Corresponding parts of congruent triangles are equal)
Hence, proved that XA = YC.
Answered By
3 Likes
Related Questions
In the given figure, ABCD is a square and P, Q, R are points on AB, BC and CD respectively such that AP = BQ = CR and ∠PQR = 90°. Prove that:
(i) PB = QC
(ii) PQ = QR
(iii) ∠QPR = 45°
In the given figure, ABCD is a square, EF || BD and R is the mid-point of EF. Prove that:
(i) BE = DF
(ii) AR bisects ∠BAD
(iii) If AR is produced, it will pass through C.
ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively such that AB = BE and AD = DF. Prove that △BEC ≅ △DCF.
The perpendicular bisectors of the sides of a △ABC meet at I. Prove that : IA = IB = IC.
