Question

In the figure below, LABC=LADC, AB=AD. Prove that triangleBCD is an isosceles triangle.​

Answer 1

Answer:

∠BCD=$90^o$

Step-by-step explanation:

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=$180^o$

∣ Angle sum property

⇒∠BCD=$90^o$  

Hence, ∠BCD is a isosceles triangle.​