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?

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

3 Answers 3


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 $Q$).

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

  • $\begingroup$ 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, 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: $$\frac{EQ}{AB}=\frac{DE}{DA}=\frac{CF}{CB}=\frac{QF}{AB}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.