# Parallelogram in an isosceles triangle [closed]

Given an isosceles $ABC$ triangle ($AB=AC$), and points $D,E$ on sides $AC, AB$, respectively, such that $AD<AE$. Let $D’$ be the reflection of $D$ to the bisector of segment $BC$. Furthermore let $E’$ be the intersection of side $AC$ and the line through $B$ parallel to $DE$. Segments $BE’, CE$ intersect at $M_1$, and segments $DE, D’E’$ meet at $M_2$.

Prove that quadrilateral $EM_1E’M_2$ is a paralelogram! I can’t prove it, angle chasing didn’t work for me. Please help!!

Extra problem: Find the angles of the parallelogram, if $\angle DBE’=\angle ECB, \angle D’CE=\angle E’BC$.

## closed as off-topic by uniquesolution, Adrian Keister, Shaun, Shailesh, NamasteAug 22 '18 at 10:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – uniquesolution, Adrian Keister, Shaun, Shailesh, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

• A picture would help! – herb steinberg Aug 21 '18 at 18:10
• (1) Through E draw a line parallel to BC. (2) Use "pairs of opposite angles are equal" to prove it is a parallelogram. – Mick Aug 21 '18 at 18:43
• Can you write a solution please! At the moment I can’t make a figure but as soon as possible I will! – Leo Gardner Aug 21 '18 at 18:55

In the following picture we have to show $EC\| D'E'$. (One pair of sides of $EM_1E'M_2$ is given as parallel. In the picture, the colored segments are given as parallel. The other pair involves the "complicated" points $M_1,M_2$, so we rewrite equivalently as above the needed parallelism. Now also try to omit the points $M_{1,2}$, and the segments connecting them to... from the above figure.)

For this, we use the converse of the theorem of Thales, here is the proof of (1) then: $$\frac{AD'}{AE} \overset{(a)}{=\!=} \frac{AD}{AE} \overset{(b)}{=\!=} \frac{AE'}{AB} \overset{(c)}{=\!=} \frac{AE'}{AC} \ .$$ So $\Delta AD'E'\sim \Delta AEC$ (proportionalities of sides building the common angle in $A$), so $D'E'\|EC$.

Comments to the above series of equalities.

• (a) uses $DD'\| BC$, so $\Delta ADD'$ is isosceles.
• (b) follows from Thales for the parallels $ED\|BE'$, or from the similitude $\Delta AED\sim \Delta ABE'$.
• (c) uses $\Delta ABC$ isosceles.

The bonus problem now. I cannot understand the given data. In the picture, the given congruent angles (and some other congruent ones) were marked in the same color. We get inscriptible (concyclic) quadrilaterals like $BCDE$ (because of the green angles in $C,D$) and/or $CDD'E$ (because of the orange angles in $C,D$).

But $BCDD'$ is an isosceles trapez, so $E$ must coincide with $D'$, contradiction to $AD<AE$.

Note: For this reason, a geometry problem is always welcome with its picture, and the picture shows often the effort already done in solving the problem. (Potential helpers can then give the proof starting from the already done steps, e.g. marked equal angles or sides, etc. - this is didactically of high benefit!)