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The problem is as follows:

The figure from below shows a parallelogram $ABCD$ and $\triangle\,APD$ and $QR=3\,cm\,RD=4\,cm$. Using this information find the value of PQ.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\frac{3}{4}\,cm\\ 2.&\frac{7}{3}\,cm\\ 3.&\frac{10}{3}\,cm\\ 4.&\frac{7}{10}\,cm\\ \end{array}$

So far the only relationship which I was able to spot was that since there's a parallelogram involved this meant.

$\angle BCA = \angle DAC.$

But that's where I'm still stuck. Is there any other relationship which can be obtained from what is given?.

Does this problem require construction or similarity or something like that?.

Please include some drawing in the answer. How, exactly, relying only in Euclidean geometry postulates can this be solved?.

amWhy
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Chris Steinbeck Bell
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1 Answers1

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Triangles $\triangle QRC$ and $\triangle DRA$ are similar to each other by A.A.A

$$\frac{QC}{AD}=\frac{QR}{RD}=\frac{3}{4}$$

Hence

$$\frac{BQ}{AD}=\frac{1}{4}$$

And triangles $\triangle PBQ$ and $\triangle PAD$ are similar to each other by A.A.A

$$\frac{PQ}{PD}=\frac{PQ}{PQ+7}=\frac{BQ}{AD}=\frac{1}{4}$$

$$4PQ=7+PQ$$

$$PQ=\frac{7}{3}$$

Lion Heart
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  • Howdy, interesting solution, but it would had been nicer (as the OP of this question) if you could have added a drawing or perhaps reusing mine to explain over the figure the places were are the similarity. As I mentioned in the question, for me to read just plain equations is not very helpful in euclidean geometry explanations. – Chris Steinbeck Bell Oct 29 '20 at 00:40
  • Please explain why $\frac{BQ}{AD}=\frac{1}{4}$ I see it many times and I just can't find a reason or a way to explain why or how did you come to that conclusion?.I'm assuming that this is due it mentions that the figure is a paralellogram and $AD=BC$ is this the reason why?. Can you better explain this part?. – Chris Steinbeck Bell Oct 29 '20 at 00:51
  • @Chris Steinbeck Bell: $AD=BC$(Opposite sides of parallelogram are equal). Since $\frac{QC}{AD}=\frac{3}{4}$, let $AD=4a$ then $QC=3a$ and $BQ=AD-QC=4a-3a=a$. Finally $\frac{BQ}{AD}=\frac{1}{4}$ – Lion Heart Oct 29 '20 at 07:31
  • @Chris Steinbeck Bell: $\triangle QRC\sim \triangle DRA $(AAA Similarity Theorem).Since $\angle CQR= \angle ADR$ and $\angle QCR= \angle DAR$ (alternate interior angles, $AD\parallel BC$).$\angle QRC=\angle ARD $(vertically opposite angles) – Lion Heart Oct 29 '20 at 07:47
  • @Chris Steinbeck Bell: $\triangle PBQ\sim \triangle PAD $(AAA Similarity Theorem).Since $\angle PBQ= \angle PAD$ and $\angle PQB= \angle PDA$ (corresponding angles, $AD\parallel BC$).$\angle BPQ=\angle APD $(common angle) – Lion Heart Oct 29 '20 at 07:53