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.


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 $|BD| = |CD|$


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 $|BD| = |CD|$

  • $\begingroup$ 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. $\endgroup$ – Rajath Krishna R Sep 1 '13 at 0:18
  • $\begingroup$ I had a little mistake, a typo that mislead you. Anyway here's a picture: img716.imageshack.us/img716/4466/vi9f.png $\endgroup$ – Stefan4024 Sep 1 '13 at 0:30
  • $\begingroup$ Wonderful.::::::::::::::::::::::::::::::: $\endgroup$ – Yadnarav3 Sep 1 '13 at 3:09
  • $\begingroup$ @Yadnarav3 Here's another solution, in my opinio even better and simplies. You can take a look at it. $\endgroup$ – Stefan4024 Sep 1 '13 at 11:59

Tip: triangles ABD, CED and ACD are all similar by AA postulate (and ultimately congruent, but that's the proof) Now what do triangles ACD and ABD have in common?

  • $\begingroup$ Could you elaborate further? $\endgroup$ – Yadnarav3 Sep 1 '13 at 3:10
  • $\begingroup$ The proof is quite easy. You state that angle CDE is right. That means triangles ABD, ACD and CED are all right. Line AD is an angle bisector as you stated and further you state that angle AEC is congruent to angle ABC. This means that triangles ABD, ACD and CED all share common angles by the Angle-Angle postulate. (If two triangles share two pair of common angle, then their third pair has to be the same too). But triangles ABD and ACD have a common side: AD, so these triangles are congruent. That implies BD=CD because these sides are across the same angle. $\endgroup$ – imranfat Sep 1 '13 at 3:27
  • $\begingroup$ Yes, but how does CDE being right prove ABD, AcD, CED are right? The angle bisect does not necessarily cross BC at D.If we call the point it does cross R, then All we could immediately say is that ABR, ACE, and REC are similar, according to my knowledge. $\endgroup$ – Yadnarav3 Sep 1 '13 at 20:11
  • $\begingroup$ Honestly, in my drawing I made the assumption that A,D and E are collinear, with D being on BC That gives 4 right angles at D. If that collinearity cannot be assumed, then I truly wonder how BD can be congruent to DC without triangle congruency. Actually I may have misunderstood the question then. $\endgroup$ – imranfat Sep 1 '13 at 23:11
  • $\begingroup$ @Rajath, I will join the club with wrong drawings :) $\endgroup$ – imranfat Sep 1 '13 at 23:12

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