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

Question

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 $\angle A$.

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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.

Answer 1

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

Answer 2

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.