AUT_ABooklet_2023

Question

Let ABCD be a rhombus with BAD < 90^. The circle passing through D with center A intersects the line CD a second time in point E. Let S be the intersection of the lines BE and AC. Prove that the points A, S, D and E lie on a circle. FIGURE_0

Accepted Answer

By the inscribed angle theorem, it is enough to show that

∠ S E D = ∠ S A D

. Since

A B C D

is a rhombus, we have

∠ S A D = \frac{1}{2} ∠ B A D .

Since

A B C E

is an isosceles trapezoid, we have by symmetry that

∠ S E D = ∠ E C S = \frac{1}{2} ∠ D C B = \frac{1}{2} ∠ B A D,

which finishes the proof.

(Theresia Eisenkölbl) □