Question

13. 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)
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Answer 1

Answer:

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Answer 2

Answer:

If in a Square ABCD

bisector of <BAC cut BD at X & BC at Y then AACY

= AABX

Step-by-step explanation:

in a square Diagonal are angle bisectors

=> <BAC = 2ACB = 90 ° / 2 = 45 °

also ZABD = 45°

Lets compare A ACY & AABX

A ACY

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

ZACY ZACB = 45° (as Y lies on BC)

ZCYA = 180 ° -45 ° -22.5 ° = 112.5 °

AABX

ZXAB=(1/2)<BAC = 22.5° ZABX = ZABD = 45° (as X lies on BD) ZAXB = 180°-45°-22.5° = 112.5°

now

2 CAY = 2XAB = 22.5°

ZACY ZABX

= 45°

ZCYA = ZAXB = 112.5 ° => AACY = AABX

AACY & AABX are similar.

Step-by-step explanation:

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