Question

In ∆ABC, AB = AC. Side BA is produced to D such that AD = AB. Prove that ∠BCD = 90°.

Answer 1

In triangle ABC, we have:

$$\begin{gathered} \mathrm{AB}=\mathrm{AC}\left(\mathrm{Given}\right) \\ \Rightarrow \angle \mathrm{ACB}=\angle \mathrm{ABC}(\mathrm{Angles}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{are}\mathrm{also}\mathrm{equal}) \end{gathered}$$
In triangle ACD, we have:
$$\begin{gathered} \mathrm{AC}=\mathrm{AD} \\ \Rightarrow \angle \mathrm{ADC}=\angle \mathrm{ACD}(\mathrm{Angles}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{side}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{are}\mathrm{also}\mathrm{equal}) \end{gathered}$$
In triangle BCD, we have:
$$\begin{gathered} \angle ABC+\angle BCD+\angle ADC=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}\mathrm{of}\mathrm{triangle}) \\ \Rightarrow \angle ACB+\angle ACB+\angle ACD+ACD=180° \\ \Rightarrow 2(\angle ACB+\angle ACD)=180° \\ \Rightarrow 2\left(\angle BCD\right)=180° \\ \therefore \angle BCD=90° \end{gathered}$$
Hence proved.