Question

In triangle ABC, AB=AC side BA is produced to D such that AD =AB prove that angle BCD =90 degree ​

Answer 1

Answer:

Step-by-step explanation:

Given: ∆ABC is an isosceles ∆.

AB = AC and AD = AB

To Prove:

∠BCD is a right angle.

Proof:

In ΔABC,

AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)

In ΔACD,

AD = AB

⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)

Now,

In ΔABC,

∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — (i)

Similarly in ΔADC,

∠CAD = 180° – 2∠ACD — (ii)

also,

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii)

∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB – 2∠ACD

⇒ 2∠ACB + 2∠ACD= 360-180

⇒ 2(∠ACB + ∠ACD) = 180°

⇒ ∠BCD = 90°

HENCE PROVED