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The assignment is as follows. Given the isosceles triangle $\triangle ABC$, on the line $AC$ a point is selected, lets call it $D$, also on the line $CB$ a point is selected, point $E$ such that $\left | AD \right |$= $\left | BE \right |$ AND points D and E are located on opposite sides of line $AB$. Proove that the line $AB$, halves the length $\overline{DE}$.

The isosceles triangle sketch for the assignment.

Basically I tried to making some sort of a parallelogram, putting down the heights in $\triangle AFD$ and $\triangle FBE$ .But nothing came of it, I was thinking along the lines of showing it is a diagonal in some parallelogram and then concluding that f is a halving point. Tried triangle simillarity, but to no avail.

I am looking for a good hint as to what I could look at to prove this, not necesarilly the entire solution.

DeLuini
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1 Answers1

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Reflect $D$ in $A$. Denote the new point by $X$. Can you prove that $XE \parallel AB$? Can you see how it follows that $F$ is the midpoint of $DE$?

timon92
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  • If I understand this right, then we get a triangle $\triangle DXE$ in which line AB is holds the triangle midsection length. Since, points $A$ and $F$ are both on the triangle, and on a line parallel to the base, it follows that $\overline{AF}$ is the triangle $\triangle DXE$ midsection, which means that point $F$ also halves $\overline{DE}$ ,but from this it follows that AB halves $\overline{DE}$. QED? – DeLuini Nov 12 '17 at 22:25
  • @DeLuini Yeah, something along these lines. Don't forget to prove in the beginning that $XE\parallel AB$! – timon92 Nov 13 '17 at 08:13
  • Thanks for the help, I've managed to prove this using your hint.I've proved first that that if a line parallel to the base goes through a side halving point that it must halve the other triangle side as well.I've proven this using the triangle congruence theorems.If anyone wants the exact details I'll write it up here. Thanks again timon92. – DeLuini Nov 19 '17 at 11:03