in the given figure ABCD and BPQ are lines. BP=BC and DQ|| CP. Prove that
a. CP = CD
b. DPbisects <CDQ

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{As} \mathrm{BP}=\mathrm{BC} \\ \Rightarrow \angle \mathrm{BPC}=\angle \mathrm{PCB} \\ \mathrm{As} \mathrm{exterior} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{opposite} \mathrm{interior} \mathrm{angles}. \\ \Rightarrow \angle \mathrm{BPC}+\angle \mathrm{PCB}=4\mathrm{x} \\ \Rightarrow \angle \mathrm{BPC}+\angle \mathrm{BPC}=4\mathrm{x} \\ \Rightarrow 2\angle \mathrm{BPC}=4\mathrm{x} \\ \Rightarrow \angle \mathrm{BPC}=2\mathrm{x}=\angle \mathrm{PCB} \\ \mathrm{Now} \mathrm{DQ}\parallel \mathrm{CP} \\ \angle \mathrm{QDC}=\angle \mathrm{PCB} \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \Rightarrow \angle \mathrm{QDC}=2\mathrm{x} \\ \mathrm{As} \mathrm{exterior} \mathrm{angle} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{opposite} \mathrm{interior} \mathrm{angles}. \\ \Rightarrow \angle \mathrm{PCB}=\angle \mathrm{CPD}+\angle \mathrm{PDC} \\ \Rightarrow 2\mathrm{x}=\mathrm{x}+\angle \mathrm{PDC} \\ \Rightarrow \angle \mathrm{PDC}=\mathrm{x} \\ \mathrm{As} \angle \mathrm{QDC}=2\mathrm{x} \mathrm{and} \angle \mathrm{PDC}=\mathrm{x} \\ \mathbf{\Rightarrow }\mathbf{DP}\mathbf{ }\mathbf{bisects}\mathbf{ }\mathbf{\angle }\mathbf{CDQ}\mathbf{ }\left[\mathbf{Hence}\mathbf{ }\mathbf{Proved}\right] \\ \mathrm{As} \angle \mathrm{PDC}=\mathrm{x} \mathrm{and} \angle \mathrm{CPD} \mathrm{and} \mathrm{since} \mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles} \mathrm{are} \mathrm{equal}, \mathrm{hence} \\ \mathbf{\Rightarrow }\mathbf{CP}\mathbf{=}\mathbf{CD}\mathbf{ }\left[\mathbf{Hence}\mathbf{ }\mathbf{Proved}\right] \end{gathered}$$

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