in given figure , AP is the bisector of angle A and PQ is the bisector of angle ACD, prove that angle APC = I/2 angle ABC.

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$$\begin{gathered} \mathrm{Let} \angle \mathrm{BAC}=\mathrm{x} ,\angle \mathrm{ACD}=\mathrm{y} \mathrm{and} \angle \mathrm{ABC}=\mathrm{z} \\ \mathrm{Now} \mathrm{exterior} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{opposite} \mathrm{interior} \mathrm{angles}, \mathrm{hence} \mathrm{in} △\mathrm{ABC} \\ \Rightarrow \angle \mathrm{ACD}=\angle \mathrm{BAC}+ \angle \mathrm{ABC} \\ \Rightarrow \mathrm{y}=\mathrm{x}+\mathrm{z} \\ \Rightarrow \mathrm{z}=\mathrm{y}-\mathrm{x} ;\left(\mathrm{i}\right) \\ \mathrm{As} \mathrm{PQ} \mathrm{is} \mathrm{bisector} \mathrm{of} \angle \mathrm{ACD} \\ \Rightarrow \angle \mathrm{ACQ}=\frac{\angle \mathrm{ACD}}{2} \\ \Rightarrow \angle \mathrm{ACQ}=\frac{\mathrm{y}}{2} \\ \mathrm{As} \mathrm{AP} \mathrm{is} \mathrm{bisector} \mathrm{of} \angle \mathrm{BAC} \\ \Rightarrow \angle \mathrm{PAC}=\frac{\angle \mathrm{BAC}}{2} \\ \Rightarrow \angle \mathrm{PAC}=\frac{\mathrm{x}}{2} \\ \mathrm{Again} \mathrm{in} ∆\mathrm{PAC} \\ \angle \mathrm{ACQ}=\angle \mathrm{PAC}+\angle \mathrm{APC} \\ \Rightarrow \frac{\mathrm{y}}{2}=\frac{\mathrm{x}}{2}+\angle \mathrm{APC} \\ \Rightarrow \angle \mathrm{APC}=\frac{\mathrm{y}-\mathrm{x}}{2} \\ \Rightarrow \angle \mathrm{APC}=\frac{\mathrm{z}}{2} ;\left[\mathrm{Using} \left(\mathrm{i}\right)\right] \\ \mathbf{\Rightarrow }\mathbf{\angle }\mathbf{APC}\mathbf{=}\frac{\mathbf{\angle }\mathbf{ABC}}{\mathbf{2}}\mathbf{ }\left[\mathbf{Hence}\mathbf{ }\mathbf{Proved}\right] \end{gathered}$$

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