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Question
An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55°.
Find:
- ∠DCB
- ∠CBD
Solution
In ΔABC,
AC = BC ...[Given]
∴ ∠CAB = ∠CBD ...[angles opp. to equal sides are equal]
⇒ ∠CBD = 55°
In ΔABC,
∠CBA + ∠CAB + ∠ACB = 180°
but, ∠CAB = ∠CBA = 55°
⇒ 55° + 55° + ∠ACB = 180°
⇒ ∠ACB = 180° − 110°
⇒ ∠ACB = 70°
Now,
In ΔACD and ΔBCD,
AC = BC ...[Given]
CD = CD ...[Common]
AD = BD ...[Given: CD bisects AB]
∴ ΔACD ≅ ΔBCD
⇒ ∠DCA = ∠DCB
⇒ ∠DCB = `(∠"ACB")/2`
= `(70°)/2`
⇒ ∠DCB = 35°
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