Finding the angle EDB in triangle ABC, where E is the intersection of the angle bisector of C with side AB and D is a point on BC

Question

This was a question I encountered while looking at some weekly math questions my school had hung in front of the department last week: I was unable to solve it, and now that some time has passed, I'd like to take a shot at it again. The problem goes as follows:

For a triangle ABC, E $\in$ [AB], and D $\in$ [BC]. [EC] is the angle bisector of ACB. (ADC) = 40° and (EC) = 20°. Find the value of (EDB).

And in case anyone's curious, m(ADC) would mean the measure (in degrees) of angle ADC.

So, I first tried using a diagram for this (the untouched form of which I've attached below) and set ECA (and since EC is the angle bisector, simultaneously the value of ECB) as equal to $\alpha$. From there, I simply tried to express every angle in terms of $\alpha$ so that I could end up with some kind of equation, but that didn't work out either. I also tried to see if I could find any similar triangles within the diagram (specifically, EDF and ACF) but since I couldn't be sure whether or not ED and AC were parallel, that didn't work out either.

So, right now, I'm sort of stumped. I don't really want a clear cut answer here: just some push in the right direction, or if there's some property here that I need to use, an explanation of said property. Trigonometry's always been one of my weaker subjects, so it seems likely there's something about angle bisectors I don't remember- but I'm not quite sure either.

Thanks in advance! Here's the diagram: (if it doesn't work, do inform me- I'll try to fix it when I can)

https://i.imgur.com/7cwFG7D.jpg

Answer 1

Start with a more realistic drawing. Ask yourself how you can draw a triangle with the desired properties. Notice in particular that $\angle ADC \cong 2\cdot \angle AEC$, and that they insisist on the same segment $AC$. This should make you think of the Central Angle Theorem.

As in the Figure above, draw the isosceles triangle $\triangle AOC$, with $\measuredangle AOC = 40^\circ$. Then $D$ lies on the circle $(AOC)$, and $E$ on the circle centered in $O$ and through $A$ and $C$.

Now concentrate on the chord $AD$. What can you say about $\measuredangle AOD$? And what about $AE$ and $\measuredangle AOE$?

We have that $\measuredangle AOD = \measuredangle ACD$. Why? Also $\measuredangle AOE = 2\cdot \measuredangle ACE = \measuredangle ACD$. (For this apply again Central Angle Theorem!)

What can you state, now, about $E$, $D$, and $O$?

They are aligned.

Observe now that $\angle ODC$ subtends the chord $OC$. We are almost done!

From the above observations, in fact $\measuredangle ODC = \measuredangle OAC = 70^\circ$, and, since $\angle ODC$ and $\angle EDB$ are vertical angles we have $$\boxed{\measuredangle EDB = 70^\circ}.$$

Answer 2

The problem is a bit tricky but is solvable without trigonometry.

Since $\angle ADC=2\angle AEC$, then $E$ must lie on the circumference of a circle with center $D$ and chord $AC$, and hence points $D$, $E$ lie not on segments $BC$, $BA$ but on their extensions, as in the figure below.

Since $\triangle ADC$ isosceles, then $\angle ACD=70^o$.

And since $CE$ bisects $\angle ACD$, and $\triangle CDE$ is isosceles, then$$\angle ECD=\angle DEC=35^o$$Therefore$$\angle EDB=180^o-70^o=110^o$$

Answer 3

This solution is probably too late in the game to be of much use, but it's worth looking at because it relies only on elementary methods such as similar triangles.

The figure includes the angle bisectors of angles $\angle ACD$, $\angle CDA$ and $\angle DAC$. We let $\alpha=\angle ACQ, \angle DCQ$ and $\beta=\angle CAF, \angle DAF$.

Note that $2\alpha+2\beta+40=180$, so that $\alpha+\beta=70$. Then, $\angle APC=180-\alpha-\beta=110$, so that $\angle APQ=70$.

Next, note that $\triangle AQE\sim\triangle PQD$, because they share two pairs of congruent angles. From this we know that $\overline{QD}/\overline{QP}=\overline{QE}/\overline{QA}$ and we know that $\angle PQA=\angle DQE$ because they are vertical angles. So, by SAS, $\triangle DQE\sim\triangle AQP$.

Thus, we know that $\angle EDQ=\angle APQ=70$. Therefore $\angle BDE=180-70-20-20=70$.