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What is the way to solve this geometry problem?

pirsquare
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1 Answers1

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Here is a hint: Suppose we draw $D$ such that $ADCB$ is a rectangle. Then extend $QP$ to $E$ on $DC$. Show that $E$ is the midpoint of $DC$, thus $\angle AQP = \angle CQB$.

heropup
  • 135,869
  • Thanks! Will give it a go! – pirsquare May 19 '14 at 09:52
  • Ok. I figured out that E is the midpoint of DC as Triangles AQP and CEP are similar. But I still don't see how this leads to angles AQP and CQB being equal. – pirsquare May 19 '14 at 14:38
  • If $E$ is the midpoint of $DC$, note that $Q$ is $1/4$ the distance from $B$ to $A$; i.e., it is the midpoint of $FB$, where $F$ is the midpoint of $AB$. – heropup May 19 '14 at 16:06
  • @heropup Can you elaborate on how did you figure out that this problem can be solved the way you did it? How did you come to that particular way of solving it? I understand all the steps, but I don't understand how people come up with solutions like these. I wouldn't have thought that this is the way to solve it. Or perhaps could you recommend some resources from which I could learn how to solve such problems myself? – BarbaraKwarc Aug 06 '16 at 14:52