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Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

Geometry Difficulty 4.9 AIME Prove it Ukraine

The equal segments $AB$ and $CD$ intersect in the point $O$ and are divided into the ratio $AO:OB=CO:OD=1:2$ . The lines $AD$ and $BC$ intersect in the point $M$ . Prove that $DM = MB$ .

Solution

Let the length of the segments be $AB=CD=3a$ , then $AO=CO=a$ and $OB=OD=2a$ . Since $\angle AOD=\angle COB$ as vertical (fig/ 17), then $\triangle AOD=\triangle COB$ . Hence $\angle ADO=\angle CBO$ . As $\triangle BOD$ is isosceles, then $\angle BDO=\angle DBO$ , consequently $\angle MDB=\angle MBD$ as the sums of the equal angles. In other words $\triangle MDB$ is isosceles, from where the equality we need.

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