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$P$ 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!

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

enter image description here

  • $\angle QBP+\angle PBA=\angle ABQ=\angle QBC+\angle ABC$ 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.

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  • $\begingroup$ Please show the diagram! $\endgroup$ – Shreya Srivastava May 23 '19 at 16:18
  • $\begingroup$ Diagram is not really helpful in this case, but added at your request anyway. $\endgroup$ – user10354138 May 23 '19 at 18:40

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