Question

Triangle ABC is an isosceles triangle in which AB=AC.
Side BA is produced to D such that AD = AB
(see Fig. 7.34). Show that angle BCD is a right angle.

​

Answer 1

Answer: ∠BCD is a right angle.

Step-by-step explanation:

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.

Answer 2

Given :-

• AB = AC

• AD = AB

That is AC = AB = AD

To Prove :-

ΔBCD = 90°

Proof :-

In ΔABC ,

AB = AC

Therefore,

ΔABC = ΔACB

[ Angles opposite to equal sides are equal ]

Now,

In ΔACD

AC = AD

ΔADC = ΔACD

[ Angles Opposite to equal sides are equal ]

Now,

In ΔBCD

ΔABC + ΔBCD + ΔBDC = 180° eq( 1 )

[ Angle sum property ]

ΔACB + ΔBCD + ΔACD = 180° eq( 2 )

[ Angle Sum Property ]

Therefore,

From ( 1 ) and ( 2 )

( ΔACB + ΔACD) + ΔBCD = 180°

ΔBCD + ΔBCD = 180°

2ΔBCD = 180°

ΔBCD = 180° / 2

ΔBCD = 90°

Hence, Proved $.$