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I have the following triangle:

enter image description here

The following information about it are given:

  • ABCD is a trapezoid (AB || DC)
  • EF || DC
  • Q is the intersection of AC, DB, PN, & EF

Prove that EQ = QF.

Since I don't have any numerical values, I tried solving it by various triangular relation identities via similarities and Thales's theorem. The only way to create a relation between EQ and QF that I could think of was this:

$$\bigtriangleup \text{APM} \sim \bigtriangleup \text{EPQ} \text{ and } \bigtriangleup \text{PMB} \sim \bigtriangleup \text{PQF}$$

$$\begin{cases} \frac{AM}{EQ} = \frac{PM}{PQ} \\ \frac{MB}{QF} = \frac{PM}{PQ} \end{cases}$$

I've then tried to swap around the redundant lengths to try and get to the desired equation, but because I lack direction and methodology I get lost and frustrated. I feel like I'm just doing guesswork.

How can I solve this particular problem, and how do I tackle problems of this kind more effectively?

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  • $\begingroup$ It is not clear to me what the definition of $Q$ is. I mean, it should be given as the intersection of only two segments. Besides, are you given information on $N$ (middle point of $DC$?) or $M$ (middle point of $AB$?)? $\endgroup$ – dfnu Jan 2 '19 at 20:46
  • $\begingroup$ @Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem $\endgroup$ – Aqua Jan 2 '19 at 20:51
  • $\begingroup$ Q is just a point. No information about middle points is given. $\endgroup$ – daedsidog Jan 2 '19 at 20:51
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First and last equality are because of triangle similarty ($BQF\sim BDC$ and $AEQ \sim ADC$) and in the middle because of Thales theorem (in angle through $Q$).

$$ {QF \over CD} = {QB\over DB} = {QA\over AC} = {EQ\over CD}$$

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  • $\begingroup$ Brilliant, thank you. No matter how many hours I gawk at these sort of problems, the obvious never occurs to me. Do you have any suggestions to remedy this? $\endgroup$ – daedsidog Jan 2 '19 at 21:01
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Let’s denote $Q$ the intersection of $AC$ and $BD$, $M$ the midpoint of $AB$ and $N$ the midpoint of $DC$.

Let’s first prove that $P, M, Q, N$ are aligned.

The homothetic transformation of center $P$ that transforms $A$ into $D$, transforms $B$ into $C$ as $ABCD$ is a trapezoid. Hence by this homothetic transformation, $M$ and $N$ are aligned with the center $P$.

Considering now the homothetic transformation of center $Q$ that transforms $A$ into $C$ and $B$ into $D$, you get by a similar argument that $M, Q,N$ are aligned.

Finally, $P, M, Q, N$ are aligned. Now, $Q$ is the middle of $EF$ by Thales theorem.

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By similarity twice and by Thales for $\angle DPC$ we obtain: $$\frac{EQ}{AB}=\frac{DE}{DA}=\frac{CF}{CB}=\frac{QF}{AB}.$$

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