Question

"Question 6 ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.

Class 9 - Math - Triangles Page 124"

Answer 1

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°

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Hope this will help you...

Answer 2

Hello !!✋

Here is your answer.

Given , ∆ABC is an isosceles triangle in which AB = AC.Side BA is produced to D such that AD = AB.

Prof that, < BCD is a right angle.

prof:-
_____

°.° ABC is a isosceles triangle

.°. < ABC = < ACB -------(1)

°.° AB = AC and AD = AB

.°. AC = AD

.°.In ∆ACD

< CDA = < ACD ( angle opposite to the equal sides are aqual )

=> < CDB = < ACD ----------(2)

From (1) and (2) we get

< ABC + < CDB = < ACB + < ACD

=> < ABC + < CDB = < BCD --------(3)

In ∆BCD,

< BCD + < DBC + < CDB = 180°
=> < BCD + < ABC + < CDB = 180°
=> < BCD + < BCD = 180° [ using (3) ]
=> 2< BCD = 180°
=> < BCD = 180°/ 2 = 90°
.°. < BCD is a right angle.

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