# Find missing angle in triangle

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.

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

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.

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 $$DE then all inequalities get reversed and we have a contradiction again. Hence $$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 $$a=10^\circ$$.

• 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 $\angle ABC + C/2=8a+a=90$ Dec 23, 2020 at 11:15
• @cosmo5 That's true, there's no need to introduce the point $G$. Thanks. Dec 23, 2020 at 11:19
• Isn't there any more elegant way to show that $AD=DE$? Anything with circles or congruent triangles? Dec 23, 2020 at 11:51
• @Samuel I just found a solution without use of inequalities. See my updated answer. Dec 23, 2020 at 13:15
• Very nice solution! Also must remark : your hand-drawn circles are good! :) 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.

• Welcome to Math.SE! Please use MathJax. For some basic information about writing math at this site see e.g. basic help on MathJax notation, MathJax tutorial and quick reference, main meta site math tutorial and equation editing how-to. Dec 23, 2020 at 8:54
• Yugikaiba, which angle are you referring to? We don't know any of the angles. If you want, you can provide an analytical solution. Dec 23, 2020 at 9:07
• @yugikaiba I found same partial solution as you've written, but angle still needs to be found, no? Dec 23, 2020 at 9:10
• In this triangle, we are given that AB=AD=CE, BE=CD AND ANGLE ECD=2∗AED=2a. Dec 23, 2020 at 9:10
• A good starting point would be to prove that $AD=DE$ (which holds). Then angle $DAE = a$ and $CDE = 2a$. Dec 23, 2020 at 10:41

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 $$a=10^{\circ}$$