# To Prove that two given triangles are similar if some conditions are given

$$P$$ is any point within triangle $$ABC$$. $$Q$$ is a point outside triangle $$ABC$$ such that $$\angle CBQ = \angle ABP$$ and $$\angle BCQ = \angle BAP$$ . Show that the triangles $$PBQ$$ and $$ABC$$ are similar. Sir here I think that the diagram I formed is not according to question!

The picture is that $$BC$$ divides $$\angle PBQ$$ internally. Then:

• $$\angle QBP+\angle PBA=\angle ABQ=\angle QBC+\angle ABC$$ and $$\angle CBQ=\angle ABP$$ gives $$\angle QBP=\angle ABC$$.

• From $$\angle CBQ=\angle ABP$$ and $$\angle BCQ=\angle BAP$$, you have $$\triangle CBQ\sim\triangle ABP$$, so $$BQ/BP=BC/AB$$.

So $$\triangle PBQ$$ and $$\triangle ABC$$ are similar (same angle and proportional sides).

Edit: The thought process that goes with this solution is as follows. If the conclusion is true, then either $$CB$$ or $$AB$$ must divide $$\angle PBQ$$. Since the given conditions on angles involve only $$\triangle CBQ$$ and $$\triangle ABP$$, it is immediate that the two are similar. Then the correct mental picture emerges --- you rotate the triangle $$ABC$$ about point $$B$$ so ray $$BC$$ becomes ray $$BQ$$. Hence the opening statement "$$BC$$ divides $$\angle PBQ$$ internally". By then there is no need to draw any any pictures.

• Please show the diagram! May 23, 2019 at 16:18
• Diagram is not really helpful in this case, but added at your request anyway. May 23, 2019 at 18:40