From triangle ABC;
[tex]\angle A+\angle B=\angle ACD\ldots\ldots.\ldots\ldots\text{.}\mathrm{}\text{equation 1}[/tex]
Also, from triangle BPC;
[tex]\begin{gathered} \angle PBC+\angle BPC=\angle PCD \\ \frac{\angle B}{2}+\angle BPC=\frac{\angle ACD}{2}\ldots\ldots...\ldots\ldots\ldots\ldots\text{.equation 2} \end{gathered}[/tex]
Then, we multiply equation 2 by 2, we have;
[tex]\angle B+2(\angle BPC)=\angle ACD\ldots\ldots.\ldots\ldots.\ldots..equation\text{ 3}[/tex]
Then, we find the difference of equation 1 and equation 3, we have;
[tex]\begin{gathered} \angle A+\angle B-\angle B-2(\angle BPC)=\angle ACD-\angle ACD \\ \angle A-2(\angle BPC)=0 \\ \angle A=2(\angle BPC) \end{gathered}[/tex]