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Bulgarian Spring Tournament

Bulgaria geometry

Problem

Given is a triangle

A B C

and two points

D \in A C

,

E \in B D

such that

∠ D A E = ∠ A E D = ∠ A B C

. Show that

B E = 2 C D

iff

∠ A C B = 9 0^{\circ}

. (Nikolay Nikolov)

Solution

Let

α = ∠ A

,

β = ∠ B

and

γ = ∠ C

. Then

\frac{B E}{sin ( α - β )} = \frac{A B}{sin ( π - β )}, \frac{C D}{sin ( α - β )} = \frac{B C}{sin 2 β}, \frac{A B}{sin γ} = \frac{B C}{sin α}

and so

\frac{B E}{C D} = \frac{sin 2 β sin γ}{sin β sin α} = \frac{2 cos β sin γ}{sin α} = \frac{2 cos β sin γ}{sin β cos γ + cos β sin γ}

Therefore

B E = 2 C D \Leftrightarrow cos γ = 0 \Leftrightarrow γ = 9 0^{\circ}

. □

Techniques

Triangle trigonometryAngle chasing