Question

3. In a square ABCD, the bisector of the angle BAC
cuts BD at X and BC at Y. Prove that the triangles
ACY, ABX are similar.
(SC)







please answer me fast ​

Answer 1

If in a Square ABCD bisector of ∠BAC  cut BD at X & BC at Y then ΔACY ≈ ΔABX

Step-by-step explanation:

in a square Diagonal are angle bisectors

=> ∠BAC  =∠ACB = 90°/2 = 45°

also ∠ABD = 45°

Lets compare Δ ACY & ΔABX

Δ ACY

∠ CAY = (1/2) ∠ BAC  = (1/2) 45° = 22.5°

∠ACY = ∠ACB = 45°  ( as Y lies on BC)

∠CYA = 180° - 45° - 22.5° = 112.5°

ΔABX

∠XAB = (1/2)∠BAC = 22.5°

∠ABX = ∠ABD  = 45°  ( as X lies on BD)

∠AXB = 180° - 45° - 22.5° = 112.5°

now

∠ CAY = ∠XAB = 22.5°

∠ACY = ∠ABX = 45°

∠CYA = ∠AXB = 112.5°

=> ΔACY ≈ ΔABX

ΔACY &  ΔABX are similar.

Similar Links :

ABCD is a square. Diagonals AC and BD intersect each other at O. Bisector of∠BAC meet BO at P and BC at Q. Prove that OP = ½ CQ

https://brainly.in/question/1114521