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In this triangle, we are given that $AB=AD=CE$, $BE=CD$ and angle $ECD=2*AED=2a$.

We are asked to find $AED=a$.

I have seen several problems similar to this but can't follow any of the solution methods. The only thing which is obvious to me is that since $AD+CD = BE+EC$, the triangle is isosceles.

Therefore its base angles (the ones I have drawn in green) are equal. Also triangle ABD is isosceles, therefore its base angles are equal.

I don't see any other relation to use.

(FYI this is not homework or anything - just practice and self-motivation). Thank you very much.

EDIT: If this is of any help, through Geogebra I get that $a=10^{\circ}$, therefore $ECD=20^{\circ}$ and we have a 80-80-20 triangle.

  • $\begingroup$ You mean $ECD=2a$, right? $\endgroup$ Dec 23, 2020 at 8:09
  • $\begingroup$ @JuanManuelPrada yes, thank you, made an edit. $\endgroup$ Dec 23, 2020 at 8:12

4 Answers 4


Choose $X$ so that $ADEX$ is a parallelogram. Then $XE=AD=CE$ and since $XE \parallel AC$, we get $\angle XEC = 180^\circ - 2a$. It follows that $a=\angle XCE$.

On the other hand, $\angle XAE = \angle DEA = a$. Therefore $\angle XCE = \angle XAE$ which shows that $AXEC$ is cyclic. Then we see that $\angle ACX = a = \angle XCE$, so the corresponding chords are equal: $AX=XE$. Also note that $X$ lies on the bisector of angle $ACB$ and since $ACB$ is isosceles, it follows that $AX=XB$.

Then we find $AX=XB=AB$ so $AXB$ is equilateral. Clearly $\angle CBX = XAC = 2a$. Writing down the sum of angles of triangle $ABC$ leads to $2a+2\cdot(2a+60^\circ)=180^\circ$ which yields $a=10^\circ$.

Below is another solution.

First we're going to show that $DE=AD$. To do so we exclude the possibilities that $DE>AD$ and $DE<AD$.

If $DE>AD$ then $\angle EAD > \angle DEA = a$. This means that $\angle EDC = a+\angle EAD > 2a = \angle DCE$ which implies that $CE>DE$ and we have a contradiction: $CE>DE>AD=CE$. If $DE<AD$ then all inequalities get reversed and we have a contradiction again. Hence $DE=AD$.

Then let $F$ be the circumcenter of $ADE$. Then $\angle DFA = 2\angle DEA = \angle ACB$, so the isosceles triangles $ADF$ and $BAC$ are similar. They are actually congruent because $AD=AB$. Since $AD=DE$, it also follows that $DEF$ is congruent to both $ADF$ and $BAC$. We can also see that $F$ lies on $AB$ because $\angle FAD = \angle BAC$.

Let $G$ be the point on $DF$ such that $DG=BE$. Then triangles $BAE$, $ADG$ and $EDG$ are congruent. In particular, $AE=EG=GA$, so triangle $AEG$ is equilateral.

What is left to do is some angle chasing. We have $\angle EGA = 60^\circ$, so $\angle DGA = 30^\circ$ as $DF$ is the axis of symmetry of $AEG$. Then $30^\circ = \angle DGA = \angle GFA + \angle FAG = \angle ACE + \angle EAC = 2a+\angle DEA = 2a+a=3a$, so $a=10^\circ$.

  • 1
    $\begingroup$ The inequality argument is sick, something I've never seen before. Once $AD=DE$ is found, a probably quicker way would be noting $\angle BFE = 4a = \angle BEF$. So $\angle ABC + C/2=8a+a=90$ $\endgroup$
    – cosmo5
    Dec 23, 2020 at 11:15
  • 1
    $\begingroup$ @cosmo5 That's true, there's no need to introduce the point $G$. Thanks. $\endgroup$
    – timon92
    Dec 23, 2020 at 11:19
  • $\begingroup$ Isn't there any more elegant way to show that $AD=DE$? Anything with circles or congruent triangles? $\endgroup$
    – Samuel
    Dec 23, 2020 at 11:51
  • $\begingroup$ @Samuel I just found a solution without use of inequalities. See my updated answer. $\endgroup$
    – timon92
    Dec 23, 2020 at 13:15
  • 1
    $\begingroup$ Very nice solution! Also must remark : your hand-drawn circles are good! :) $\endgroup$
    – cosmo5
    Dec 23, 2020 at 13:29

Using Sine law,

Say, $\angle DAE = u, AB = x, DE = y, AE = z$

In $\displaystyle \triangle ADE, \frac{\sin u}{y} = \frac{\sin a}{x} \implies y = \frac{x \sin u}{\sin a}$

In $\displaystyle \triangle CDE, \frac{\sin 2a}{y} = \frac{\sin (a+u)}{x} \implies y = \frac{x \sin 2a}{\sin (a+u)}$

This leads to, $\, \sin u \sin(a+u) = \sin a \sin 2a$

Using $\cos(A - B) - \cos(A+B) = 2 \sin A \sin B,$ we get $u = a$.

Now in $\triangle ADE,$

$\displaystyle \frac{\sin 2a}{z} = \frac{\sin a}{x} \implies z = 2x \cos a$

In $\triangle ABE,$

$\angle ABE = 90^0 - a, \angle BAE = (90^0 - a) - a = 90^0 - 2a, \angle BEA = 3a$

$\displaystyle \frac{\sin 3a}{x} = \frac{\sin (90^0 - a)}{z}$

$\displaystyle \sin 3a = \frac{1}{2} \implies a = 10^0$


First, from Euclid: an inscribed angle is half of the central angle that subtends the same arc on the circle.

Second, if you already have the angle 2a finding the angle a is trivial with a compass you can easily find the bisector, if you want to do it "pretty", if you are allowed to replicate C (with AD as a base instead of AB) and call that point F, and draw a circunference of center F and radius AF, the intersection with BC should give you point E.


Using brute force.

Calling $u$ the segment with two marks and $v$ the segment with three marks we have

$$ \cases{ v^2=2(u+v)^2-2(u+v)^2\cos(2a)\Rightarrow a=\frac 12\arccos\left(\frac{2u^2+4u v+v^2}{2(u+v)^2}\right)\\ v^2=ED^2+AE^2-2ED\cdot AE\cos(a)\\ ED^2=u^2+v^2-2u v\cos(2a)\\ AE^2=u^2+v^2-2u v\cos(\phi)\\ \phi = \frac{\pi}{2}-a } $$

making substitutions we get at

$$ \sqrt{\frac{(2 u+v) (2 u+3 v) \left(u^3+u^2 v+v^3\right) \left(u^4-2 u^2 v^2+u v^3+v^4\right)}{(u+v)^5}}-\frac{v^3 (2 u+v)}{(u+v)^2}+2 u v-2 u^2=0 $$

making now $v = \lambda u$ we obtain

$$ \left(\sqrt{\frac{(\lambda +2) (3 \lambda +2) \left(\lambda ^3+\lambda +1\right) \left(\left(\lambda ^2+\lambda -2\right) \lambda ^2+1\right)}{(\lambda +1)^5}}-\frac{\lambda \left(\lambda ^3-2 \lambda +2\right)+2}{(\lambda +1)^2}\right) u^2=0 $$

and the $\lambda$ factor reduces to

$$ \frac{\lambda ^2 \left(2 (\lambda +2) \lambda ^3+\lambda ^2+\lambda +1\right) \left(\lambda ^2 (\lambda +3)-1\right)}{(\lambda +1)^5}=0 $$

with the only feasible root

$$ \lambda = 0.532088886237956 $$

which corresponds to $a=10^{\circ}$


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