# Can't figure out this triangle geometry problem

I have the following triangle:

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?

• 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 $AB$?)?
– dfnu
Jan 2, 2019 at 20:46
• @Matteo You need no information about $N$ and $M$ since you can deduce that are midpoint by say Ceva-theorem Jan 2, 2019 at 20:51
• Q is just a point. No information about middle points is given. Jan 2, 2019 at 20:51

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 $$Q$$).

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

• 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? Jan 2, 2019 at 21:01

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.

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