Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it Romania

Let $\triangle ABC$ be a triangle with $\angle BAC = 30^\circ$ and $\angle ABC = 100^\circ$ . Let $m$ be the perpendicular bisector of $AC$ , $E$ be the intersection of $m$ and $AB$ , and $D$ be the point on $m$ , inside the triangle $ABC$ , such that $\angle CAD = 10^\circ$ . Let $M$ be the intersection of the lines $AD$ and $CE$ .

a) Prove that $CE$ is the bisector of the angle $\angle BCD$ .

b) Prove that $AM = AB$ .

Solution

a) As $m$ is the perpendicular bisector, we have $DA = DC$ and $EA = EC$ . Thus $\triangle DEA \equiv \triangle DEC$ .

From $DA = DC$ , we get $\angle DCA = \angle DAC = 10^\circ$ . We have $\angle DCE = \angle DAE = 20^\circ$ , so $\angle BCE = 20^\circ$ and $CE$ is the bisector of the angle $BCD$ .

b) As $\angle CDA = 160^\circ$ , it follows $\angle ADE = \angle CDE = 100^\circ = \angle CBE$ .

We obtain $\angle CEB = \angle CED = 60^\circ$ , so $\triangle CEB \equiv \triangle CED$ (A.S.A.)

It follows that $BE = BD$ . But $\angle MEB = \angle MED$ and $ME = ME$ , so $\triangle MEB \equiv \triangle MED$ (S.A.S.).

We infer that $\angle EBM = \angle EDM = 180^\circ - \angle CED - \angle AME = 180^\circ - 60^\circ - 40^\circ = 80^\circ$ . But $\angle BAM = 20^\circ$ , so $AB = AM$ .

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