# Finding the sum of $5$ angles given equivalent division of a side

Let $$ABC$$ be an isosceles triangle with $$AB = AC.$$ Point $$D$$ lies on segment $$AB$$ so that $$AD = AB/6.$$ Points $$E_1, E_2, E_3, E_4,$$ and $$E_5$$ lie on segment $$BC$$, in this order from $$B$$ to $$C$$, and they divide it into six equal parts. Find $$\angle AE_1D + \angle AE_2D + \angle AE_3D + \angle AE_4D + \angle AE_5D$$ in terms of $$\angle A$$.

I have experimented with this problem using a Geogebra diagram and concluded that the answer was $$\angle A / 2$$, but I am completely unsure how to start on this problem since I can't find any useful common angles.

Note that, by symmetry, $$\angle AE_1C+\angle AE_5C=\pi$$ $$\angle AE_2C+\angle AE_4C=\pi$$ $$\angle AE_3C=\frac{\pi}{2}$$

Therefore $$\sum \angle AE_iC=\frac{5\pi}{2}.$$

Now notice that triangle $$DBE_5$$ is similar to triangle $$ABC$$. By the same argument as above we have:

$$\sum \angle DE_iC=\angle DE_5C+2\pi=3\pi-\angle C.$$

Thus $$\sum \angle AE_iD=\sum \angle DE_iC-\sum \angle AE_iC=\frac{\pi}{2}-\angle C=\frac{1}{2}\angle A.$$

For convenience, let $$\alpha = \angle A$$, $$\beta = \angle B = \angle C = \frac{\pi - \beta}{2}$$, $$x = AD$$, and $$y = \frac{BC}{6}$$. We want to find $$\epsilon_k = \angle AE_kD$$ for $$k \in \{1, 2, 3, 4, 5\}$$.

Set up a coordinate system with $$B = (0, 0)$$ and $$C = (6y, 0)$$, and it follows that $$A = (7x \cos \beta, 7x \sin \beta)$$, $$D = (6x \cos \beta, 6x \sin \beta)$$, and $$E_k = (ky, 0)$$. Then:

• $$DE_k = \sqrt{(6x \cos \beta - ky)^2 + (6x \sin \beta)^2} = \sqrt{36x^2 - 12kxy \cos \beta + k^2y^2}$$

• $$AE_k = \sqrt{(7x \cos \beta - ky)^2 + (7x \sin \beta)^2} = \sqrt{49x^2 - 14kxy \cos \beta + k^2y^2}$$

Knowing the lengths of all three sides of $$\triangle AE_kD$$ (the other side is just $$x$$), you can use the Law of Cosines to solve for $$\cos \epsilon_k$$. However, when I attempted to do this, I got a bunch of tedious algebra, so maybe there's an “obvious” trick to make things simpler.