Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Prove it Austria

Let ABCABC be a triangle with circumcenter UU such that ∠CBA=60∘\angle CBA = 60^\circ and ∠CBU=45∘\angle CBU = 45^\circ. Let DD be the point of intersection of the lines BUBU and ACAC.
Prove that AD=DUAD = DU.

Solution

Figure 1

In the isosceles triangle AUBAUB, we have
∠BAU=∠UBA=60∘−45∘=15∘, \angle BAU = \angle UBA = 60^\circ - 45^\circ = 15^\circ,
and therefore
∠AUB=180∘−∠BAU−∠UBA=150∘. \angle AUB = 180^\circ - \angle BAU - \angle UBA = 150^\circ.
The inscribed angle theorem implies
∠BCA=12∠BUA=75∘, \angle BCA = \frac{1}{2} \angle BUA = 75^\circ,
and therefore
∠BAC=180∘−60∘−75∘=45∘. \angle BAC = 180^\circ - 60^\circ - 75^\circ = 45^\circ.
We can finally compute the two angles of interest:
∠UAD=∠BAD−∠BAU=∠BAC−∠BAU=45∘−15∘=30∘ \angle UAD = \angle BAD - \angle BAU = \angle BAC - \angle BAU = 45^\circ - 15^\circ = 30^\circ
∠DUA=180∘−∠AUB=180∘−150∘=30∘ \angle DUA = 180^\circ - \angle AUB = 180^\circ - 150^\circ = 30^\circ
Therefore, the triangle AUDAUD is isosceles with apex DD and we have AD=DUAD = DU as desired.

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