Math, asked by rpreddy62p5s5jg, 1 year ago

A triangle ABC is right angled at A . AL is drawn perpendicular to BC . Prove that angle BAL = ACB .

Answers

Answered by alisha345
12
HERE IS YOUR ANSWER____
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Given: ABC is right triangle right angled at A and AL ⊥ BC

Now, in ΔBAL and ΔBCA

∠ABL = ∠CBA  (Common)

∠BLA = ∠BAC = 90°

So, ΔBAL ∼ ΔBCA  (by AA similarity criteria)

⇒ ∠BAL = ∠ACB  (C.P.C.T) 

@HOPE IT HELPS YOU!!

rpreddy62p5s5jg: there is no aa criteria
alisha345: sry by mistake it was written
Answered by Hardiksahu
7

No need to do any congruency stuff.

Also It is not given that BL = CL. It can be simply done,

Angle BAL = BAC - LAC

=90° - LAC (1)

In triangle ADC, by angle sum property,

ALC + LAC + ACL = 180°

90° + LAC + ACB = 180° (ACL=ACB) ( common)

ACB = 90° - LAC (2)

From (1) and (2),

BAL = ACB = 90° - LAC

Hence Proved

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