Tangents and Triangles

We consider $AX$ first. $\angle AXC = \angle ABC$ (angles in the same segment) and $\angle BXC = 180^{\circ} - \angle BAC$ (opposite angles in a cyclic quadrilateral), $\angle XBC = \angle BCA$ (alternate angles).

Now, since $\angle ABX = \angle BXC$ and $BX$ is parallel to $AC$ , we can conclude that $ABXC$ is a isosceles trapezium (or trapezoid). Hence the diagonals are of the same length so $AX = BC$ . Similar reasoning shows that $AY = BC$ .

Therefore $AX + AY = 2BC = 20$ , as required.

Cheaty way to do this :P, but since we dont need a proof to get a problem right on Brilliant, this is totally viable. WLOG, assume that ABC is equilateral. Hence, X = B and C = Y. AX = AB = AC = AY = BC = 10. AX + AY = 10 + 10 = 20

Let $a$ be the perpendicular bisector of segment $AC$ , and of $BX$ too (parallel chords). Reflection over $a$ maps triangle $\triangle ABX$ onto triangle $\triangle CBX$ , whence $|AX|=|BC|$ . Similarly $|AY| = |BC|$ and eventually $|AX| + |AY| = 2 |BC|$ . $\square$

MAIN IDEA: Supplementary angles subtended on the circumference of a circle must cut equal chords, because two supplementary angles can be considered as two opposite vertices of a cyclic quadrilateral (which are known to add up to 180°) and the common chord they cut is the diagonal passing through the other two vertices of the cyclic quadrilateral.

Here angles BAC and ACY add up to 180° hence the chords they cut viz. BC and AY must be equal.

Same on the other side for AX

Thus AX = AY = BC = 10

And AX + AY = 20