In the given figure, $$\triangle ABC$$ is an isosceles triangle, and $$AB = AC$$. $$\measuredangle BAC = 20^\circ$$ and $$BD$$ is an angle bisector of $$\measuredangle ABC$$. $$O$$ is the circumcentre of the triangle. Find the value of $$\measuredangle ADO$$.

I have been trying to solve this problem which was given to us, I have made a little progress please guide me further. $$\angle ABC = \angle ACB = 80^\circ$$ $$\angle ABD = \angle DBC = 40^\circ$$ We can also find $$\angle AOB$$ And $$\angle AOC$$ $$\angle AOB = \angle AOC = 160^\circ$$ And how can we prove that $$BD = DE = AE$$?

Thanks!

• Unfortunately, I feel I had to flag this. This isn’t a site that will solve problems for you, and some effort or examples of your own thoughts must be shown. That way, we can help you—and others—the best we can. For more information, please read how to ask a good question. At any rate, I wish the best of luck to you, and please continue to contribute to our wonderful site! – gen-ℤ ready to perish Jun 20 '20 at 18:22
• what about now? – Shyspiderman Jun 20 '20 at 18:42
• That’s much better. Having made that edit will put it in the reopen queue. – gen-ℤ ready to perish Jun 20 '20 at 18:42
• Note: $\frac{|CD|}{|DA|}=\frac{|BC|}{|AB|}$ – Invisible Jun 20 '20 at 19:06
• Circumcenter of an isosceles triangle lies on the altitude from the vertex $A$, and, it also happens to be the bisector of the $\measuredangle CAB$. Using the law of cosines, we can find $|OD|$ and then use the law of sines to find $\measuredangle ODA$. – Invisible Jun 20 '20 at 19:49

If you are looking for a pure geometric solution, the answer given below may not meet your expectations, because it contains some elements of trigonometry. We would like to emphasize here that out of all options we have explored, this particular approach demands the least effort.

The dotted lines shown in our diagram are the auxiliary lines, which we need to do the job. Let us start by finding some angles, which we are going to use later in our proof.

Since $$O$$ is the circumcentre of the isosceles triangle $$ABC$$, $$OA$$ bisects its apex angle $$\measuredangle CAB$$. Therefore, we have, $$\measuredangle DAO=10^o. \tag{1}$$

$$DG$$ is the perpendicular line dropped from $$D$$ to the baseline $$BC$$ to meet it at $$G$$. The line $$DF$$ is parallel to $$BC$$ and it meets the extended $$AO$$ at $$N$$. Furthermore, the extended $$AO$$ intersects $$BC$$ at $$M$$ and they are perpendicular to each other.

Since $$BD$$ is the angle bisector of $$\measuredangle ABC$$, $$\measuredangle FBD$$ and $$\measuredangle DBC$$ are equal to $$40^o$$. Since $$DF$$ is parallel to $$BC$$, $$\measuredangle BDF$$, which is an alternate angle to $$\measuredangle DBM$$, is also equal to $$40^o$$. This makes $$FBD$$ an isosceles triangle. Therefore, we can state, $$DF=FB .\tag{2}$$

Now, let $$FB = h$$. According to equation (2), $$DF=h$$ too. It is also evident that $$DC=FB=h \tag{3},$$ $$MG=ND=\frac{DF}{2}=\frac{h}{2}. \tag{4}$$

Consider the right-angled triangle $$DGC$$. We know that $$\measuredangle GCD=80^o$$. Therefore, we have $$\measuredangle CDG = 10^o$$. Therefore, $$GC=h\cos\left(80^o\right) \quad\mathrm{and}\quad DG=h\cos\left(10^o\right). \tag{5}$$

Using equations (4) and (5), we obtain, $$BM=MC=MG+GC=\frac{1}{2}FD+GC=\frac{h}{2}\Big(1+2\cos\left(80^o\right)\Big).$$

Since $$\measuredangle MOB$$ is an angle subtended at the circumcentre $$O$$, its magnitude can be expressed as $$\measuredangle MOB = 2\times \measuredangle BAM = 20^o$$. Considering the triangle $$BMO$$, we can write, $$OM=BM\cot\left(20^o\right)=BM\tan\left(70^o\right)= \frac{h}{2}\Big(1+2\cos\left(80^o\right)\Big)\tan\left(70^o\right). \tag{6}$$

We can use equations (5) and (6) to obtain an expression for $$ON$$ as shown below. $$ON=OM-NM=OM-DG=\frac{h}{2}\Big(1+2\cos\left(80^o\right)\Big)\tan\left(70^o\right) -h\cos\left(10^o\right) \tag{7}$$

Consider the right-angled triangle $$OND$$. Let $$\measuredangle NDO = \theta$$. Using equatioms (4) and (7), we can express $$\tan\left(\theta\right)$$ in terms of known angles. $$\tan\left(\theta\right) = \frac{ON}{ND} =\Big(1+2\cos\left(80^o\right)\Big)\tan\left(70^o\right) -2\cos\left(10^o\right)=\cot\left(20^o\right)-\sec\left(10^o\right)=1.732051$$

This means $$\measuredangle NDO = \tan^{-1}\left(1.732051\right)=60^o$$. Therefore, $$\measuredangle DON=30^o$$. Since $$\measuredangle DON$$ is an exterior angle of the triangle $$AOD$$, we have, $$\measuredangle ODA=\measuredangle DON - \measuredangle DAO =30^0-10^0 = 20^0.$$