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