Question

In Fig., ∆ABC is right angled at A. Q and R are points on line BC and P is a point such that QP || AC and RP || AB. Find ∠P.

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{given}\mathrm{triangle},\mathrm{AC}\parallel \mathrm{QP}\mathrm{and}\mathrm{BR}\mathrm{cuts}\mathrm{AC}\mathrm{and}\mathrm{QP}\mathrm{at}\mathrm{C}\mathrm{and}\mathrm{Q},\mathrm{respectively}. \\ \therefore \angle \mathrm{QCA}=\angle \mathrm{CQP}(\mathrm{Alternate}\mathrm{interior}\mathrm{angles}) \\ \mathrm{Because}\mathrm{RP}\parallel \mathrm{AB}\mathrm{and}\mathrm{BR}\mathrm{cuts}\mathrm{AB}a\mathrm{nd}\mathrm{RP}\mathrm{at}\mathrm{B}\mathrm{and}\mathrm{R},\mathrm{respectively},\angle \mathrm{ABC}=\angle \mathrm{PRQ}(\mathrm{alternate}\mathrm{interior}\mathrm{angles}). \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Hence},\mathrm{for}△\mathrm{ABC},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC}=180° \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ACB}+90°=180°(\mathrm{Right}\mathrm{angled}\mathrm{at}\mathrm{A}) \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ACB}=90° \\ \mathrm{Using}\mathrm{the}\mathrm{same}\mathrm{logic}\mathrm{for}△\mathrm{PQR},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{PQR}+\angle \mathrm{PRQ}+\angle \mathrm{QPR}=180° \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{QPR}=180°(\because \angle \mathrm{ABC}=\angle \mathrm{PRQ}\mathrm{and}\angle \mathrm{QCA}=\angle \mathrm{CQP}) \\ \mathrm{Or}, \\ 90°+\angle \mathrm{QPR}=180°(\because \angle \mathrm{ABC}+\angle \mathrm{ACB}=90°) \\ \mathbf{\angle }\mathbf{QPR}\mathbf{=}\mathbf{90}\mathbf{°} \end{gathered}$$