The midpoint of a base in a trapezoid ABCD is connected with the vertexes of another base. These lines intersect the diagonals of the trapezoid in points P and Q. How can I prove that FP = PQ = QG?
Prove that three line segments are equal.
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geometry
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0What equivalences have you established already? – 2017-02-17
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1I have found out that PQ is parallel to AD. – 2017-02-17
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0Good, then show that $PQ/AE=CQ/CE=QG/DE$ and $PQ/DE = BP/DE=FP/AE$. – 2017-02-17
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0That's because ACE and CPQ are similar, and the second one is because BPQ and BED are similar too. What now? – 2017-02-17
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0Have you forgotten what $E$ is? – 2017-02-17
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0Hint: if you erase vertex $C$ and all the segments connected to it, but you know that $\overline{PQ}$ is parallel to $\overline{AD},$ can you prove that $FP = PQ$ using only the parts of the figure that remain? (That is, using triangle $ABD,$ lines $\overline{BE}$ and $\overline{FQ},$ and the point $P$?) – 2017-02-17
1 Answers
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Sufficient hints have been given, I just complete it.
In the figure, the brown triangles are similar. This means $\dfrac {CQ}{QE} = \dfrac {BC}{ED}$.
Similarly, from the green similar triangles, we have $\dfrac {CP}{PA} = \dfrac {BC}{AE}$.
$AE = ED \rightarrow \dfrac {CQ}{QE} = \dfrac {CP}{PA}$.
The above, together with the common angle, makes $\triangle CPQ \sim \triangle CAE$. This means PQ // AD.
From that, we have $\triangle BFP \sim \triangle BAE$ and $\triangle BPQ \sim \triangle BED$.
Then, $\dfrac {FP}{AE} = \dfrac {BP}{BE} = \dfrac {PQ}{ED}$.
Therefore, FP = PQ since AE = ED.
Similarly, we have PQ = QG.