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Answer:
here you here's the 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°
Hope this will help you...
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Hii.. Good Afternoon .. I'm fine :) WBY?
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