Question

ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

​

Answer 1

Answer:

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°

Answer 2

Answer:

Answer

In △ABC, we have

AB=AC ∣ given

∠ACB=∠ABC ... (1) ∣ Since angles opp. to equal sides are equal

Now, AB=AD ∣ Given

∴AD=AC ∣ Since AB=AC

Thus , in △ADC, we have

AD=AC

⇒∠ACD=∠ADC ... (2) ∣ Since angles opp. to equal sides are equal

Adding (1) and (2) , we get

∠ACB+∠ACD=∠ABC+∠ADC

⇒∠BCD=∠ABC+∠BDC ∣ Since∠ADC=∠BDC

⇒∠BCD+∠BCD=∠ABC+∠BDC+∠BCD ∣ Adding ∠BCD on both sides

⇒2∠BCD=180

∘

∣ Angle sum property

⇒∠BCD=90

∘

Hence, ∠BCD is a right angle.