$\Delta ABC$ is right angled triangle. $AP$ and $AQ$ meet $BC$ and $BC$ produced in $P$ and $Q$ and are equally inclined to $AB$.

Question

$\Delta ABC$ is right angled triangle at $A$. $AP$ and $AQ$ meet $BC$ and $BC$ produced respectively in $P$ and $Q$ and are equally inclined to $AB$ ($\angle BAP=\angle BAQ$). Show that $\frac{BP}{BQ}=\frac{CP}{CQ}$.

I used angle bisector theorem and found that $\frac{BP}{BQ}=\frac{AP}{AQ}$. But I don't know how to proceed further. Can this be solved without trigonometry? Thanks in advance.

Answer 1

Hint: Finish off with the Sine Law.

From angle bisector theorem, $$\frac {CP}{CQ} = \frac {AP}{AQ}$$ The notations are different since I extended $BC$ towards $C$. By Sine Law, $$\frac {AP}{\sin B} = \frac{BP}{\sin \angle PAB} = \frac {BP}{\sin (90^\circ - \angle PAC)}$$ $$\frac {AQ}{\sin B} = \frac{BQ}{\sin \angle QAB} = \frac {BQ}{\sin (90^\circ + \angle QAC)} = \frac {BQ}{\sin (90^\circ - \angle PAC)}$$ Hence $$\frac {AQ}{BQ} = \frac {AP}{BP}$$$$\leadsto\frac {AP}{AQ} = \frac {BP}{BQ} = \frac {CP}{CQ}$$

Answer 2

HINT: Construct $QQ'$ with $Q'$ on $AC$ extended such that $QQ'=QA$

From angle bisector theorem, $$\frac{BP}{BQ}=\frac{AP}{AQ}$$ In $\Delta CPA$ and $\Delta CQQ'$
$$\angle BAP=\angle BAQ \Rightarrow 90-\angle BAP=90-\angle BAQ \Rightarrow \angle PAC=\angle QAQ'=\angle QQ'C$$ $$\angle C = \angle C$$ Therefore, $$\Delta CPA \sim \Delta CQQ'$$ Hence, by similarity, $$\frac{BP}{BQ}=\frac{AP}{AQ}=\frac{AP}{QQ'}=\frac{CP}{CQ}$$ Therefore, the proof is done without trigonomery.