Question

ΔABC is right angled at A and AL ┴ BC. Prove that ∠BAL = ∠ACD.

Answer 1

Answer:

Given data: ΔABC is right angled triangle at A.

AL ┴ BC

To prove: ∠BAL = ∠ACB

Proof:  

Since, ΔABC is right angled triangle at A therefore, ∠A = 90° .

In ΔABC, we have sum of all the angles of a triangle is equal to 180°. So,

∠A + ∠B + ∠C = 180°

Or, 90° + ∠B + ∠C = 180°

Or, ∠B + ∠C = 180° - 90° = 90°

Or, ∠C = 90° - ∠B

Or, ∠ACB = 90° - ∠B …… (i)

Now, considering ΔABL, we have

∠ABL + ∠BAL + ∠BLA = 180°

Since AL ┴ BC therefore, ∠BLA = 90°

∠ABL + ∠BAL + 90°= 180°

Or, ∠ABL + ∠BAL = 180° - 90°= 90°

Or, ∠BAL = 90° - ∠ABL

Or, ∠BAL = 90° - ∠B ……. (ii)

From equation (i) & (ii), we get

∠ACB = ∠BAL

Hence proved  

Answer 2

Answer:

Step-by-step explanation:

Proof:  

Since, ΔABC is right angled triangle at A therefore, ∠A = 90° .

In ΔABC, we have sum of all the angles of a triangle is equal to 180°. So,

∠A + ∠B + ∠C = 180°

Or, 90° + ∠B + ∠C = 180°

Or, ∠B + ∠C = 180° - 90° = 90°

Or, ∠C = 90° - ∠B

Or, ∠ACB = 90° - ∠B …… (i)

Now, considering ΔABL, we have

∠ABL + ∠BAL + ∠BLA = 180°

Since AL ┴ BC therefore, ∠BLA = 90°

∠ABL + ∠BAL + 90°= 180°

Or, ∠ABL + ∠BAL = 180° - 90°= 90°

Or, ∠BAL = 90° - ∠ABL

Or, ∠BAL = 90° - ∠B ……. (ii)

From equation (i) & (ii), we get

∠ACB = ∠BAL

Hence proved