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12/07/2021 11:40 am
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ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Figure). Show that BCD is a right angle.
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12/07/2021 11:41 am
It is given that AB = AC and AD = AB
We will have to now prove BCD is a right angle.
Proof:
Consider ΔABC,
AB = AC (It is given in the question)
Also, ACB = ABC (They are angles opposite to the equal sides and so, they are equal)
Now, consider ΔACD,
AD = AB
Also, ADC = ACD (They are angles opposite to the equal sides and so, they are equal)
In ΔABC,
CAB + ACB + ABC = 180°
CAB + 2ACB = 180°
⇒ CAB = 180° – 2ACB — (i)
Similarly, in ΔADC,
CAD = 180° – 2ACD — (ii)
CAB + CAD = 180° (BD is a straight line.)
Adding (i) and (ii) we get,
CAB + CAD = 180° – 2ACB + 180° – 2ACD
⇒ 180° = 360° – 2ACB-2ACD
⇒ 2(ACB+ACD) = 180°
⇒ BCD = 90°