Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.0 National olympiad, round 2 Prove it Romania

The triangle $ABC$ has $\angle BAC = 54^\circ$ and $\angle ACB = 45^\circ$ . Let $D$ and $E$ be points on the segments $BC$ and, respectively, $AD$ so that $AD = AB$ and $BE = BD$ . Denote $F$ the common point of the lines $BE$ and $AC$ , and let $DM$ be the bisector of the angle $\angle ADF$ , where $M \in AC$ . The perpendicular from $C$ on $AB$ meets $DM$ in $G$ . Prove that:

a) the triangle $ABF$ is isosceles;

b) $CG = CM$ .

Solution

a) From the statement, $\angle ABC = 81^\circ$ and, in the isosceles triangle $ABD$ , $\angle ADB = \angle ABD = 81^\circ$ , hence $\angle DAB = 18^\circ$ and $\angle CAD = \angle CAB - \angle DAB = 36^\circ$ .
The isosceles triangle $BED$ yields $\angle BED = \angle BDE = 81^\circ$ , therefore $\angle DBE = 18^\circ$ , whence $\angle ABF = \angle ABC - \angle EBD = 63^\circ$ .

From the triangle $ABF$ , $\angle ABF = 63^\circ$ and $\angle FAB = 54^\circ$ , hence $\angle AFB = 63^\circ$ , that is triangle $ABF$ is isosceles, with $AF = AB$ .

b) The relations $AD = AB$ and $AF = AB$ lead to $AF = AD$ , so the triangle $AFD$ is isosceles, with $\angle AFD = \angle ADF = \frac{180^\circ - \angle FAD}{2} = 72^\circ$ . Since $DM$ is the bisector of the angle $ADF$ , $\angle ADM = \angle FDM = \frac{\angle ADF}{2} = 36^\circ$ . The triangle $AMD$ gives $\angle MAD = \angle ADM = 36^\circ$ , therefore $\angle CMD = 72^\circ$ .
From $CG \perp AB$ follows $\angle ACG = 90^\circ - \angle CAB = 36^\circ$ , so the triangle $CMG$ has $\angle ACG = 36^\circ$ and $\angle CMG = 72^\circ$ . This gives $\angle CGM = 72^\circ$ , that is the triangle $CMG$ is isosceles, with $CG = CM$ .

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