# math contest geometry proof problem

Could someone help me with this?

Suppose $A,B,C$ are vertices of a triangle and $D$ is a point on the side $BC$. Let $l$ be the line that contains $A$ and bisects $∠CAB$. Suppose there is a point $E$ on $l$ such that upon drawing the line segments $EC$ and $DE$, we have $∠AEC = ∠ABC$ and $∠CDE = 90 ^{\circ}$. Then, show that $|BD|$ = $|CD|$. (Note: $|XY|$ denotes the length of the line-segment $XY$ .)

Drawing a circle around the figure so the base is a chord is what I tried for a while. I couldn't get anywhere else, but you may have more luck with it than I did.

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We draw the circumcircle of the triangle $ABC$. The condition states that $∠AEC = ∠ABC$, which means that the point $E$ lies on the circumcircle, beacuse the both angles cut the same chord $AC$. This means that the quadrilaterial $ABCE$ is cyclic.

We draw a normal line to the side $BC$. This line is the line $l$ and cuts the side $BC$ at the point $D$ and the circumcircle of the triangle $ABC$ at point $F$. The quadrilaterial $BCEF$ is cyclic. Here's the proof:

From the quadrialterial $ABCE$ we have $∠BEC + ∠BAC = 180^{\circ}$. Because angles $∠BAC$ and $∠BFC$ lie above the same arch they are equal so this implies $∠BEC + ∠BFC = 180^{\circ}$. And because they are opposite angle it means that $BECF$ is cyclic.

Because the diagonals in the cyclic quadrilaterial $BECF$ are normal to each other it implies that $BECF$ is square (only if ABC is right trinagle) or kite. In any case the diagonal $EF$ cut the other diagonal $BC$ in half.

This leads to $|BD| = |CD|$

Q.E.D.

Here's one eve simplier solution.

Note that for fixed point $B$ and $C$ and fixed circumcircle, the point $E$ will always be on the same position, no matter where $A$ lies on the circumcircle. This is due the fact that the bisector of the $∠BAC$ divides it into two equal angles. Both angles $∠BAE$ and $∠CAE$ are equal and are inscribe angles in the circumcircle, which implies they lie on arches and chords of the same length, i.e $|BE| = |CE|$

Now applying the Pythagorean Theorem on the right triangles $BED$ and $CED$ we have:

$$BD^2 = BE^2 + ED^2 \text{ and } CD^2 = CE^2 + ED^2 = BE^2 + ED^2$$

This leads to $BD^2 = CD^2$, which implies $|BD| = |CD|$

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if you are good in drawing figures using paint or something I would recommend a picture regarding the above problem.....I guess I have drawn the wrong way. –  Rajath Krishna R Sep 1 at 0:18
I had a little mistake, a typo that mislead you. Anyway here's a picture: img716.imageshack.us/img716/4466/vi9f.png –  Stefan4024 Sep 1 at 0:30
Wonderful.::::::::::::::::::::::::::::::: –  Yadnarav3 Sep 1 at 3:09
@Yadnarav3 Here's another solution, in my opinio even better and simplies. You can take a look at it. –  Stefan4024 Sep 1 at 11:59