Question

I need the answer quickly, I'll mark as brainliest also, please help!​

Answer 1

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

answer :--

Proof: In triangle ALB,

∠ALB + ∠BAL + ∠ABL = 180°

| Angle sum property of a triangle

⇒ 90° + ∠BAL + ∠ABC = 180°

⇒ ∠BAL + ∠ABC = 90° ...(1)

In triangle ABC,

∠BAC + ∠ACB + ∠ABC = 180°

| Angle sum property of a triangle

⇒ 90° + ∠ACB + ∠ABC = 180°

⇒ ∠ACB + ∠ABC = 90° ...(2)

From (1) and (2),

∠BAL + ∠ABC = ∠ACB + ∠ABC

⇒ ∠BAL = ∠ACB

this is your answer

Answer 2

In ΔALB

<ALB + <BAL + < ABL = 180

90 + <BAL + <ABC = 180

<BAL + <ABC = 90. ......(1)

In ΔABC

<BAC + <ABC + <ACB = 180

90 + <ACB + <ABC = 180

<ACB + <ABC = 90. ......(2)

From (1) and (2), We have

<BAL = <ACB