Ukrajina, 2008

Geometry Difficulty 5.0 AIME Prove it Ukraine

Diagonals $AC$ and $BD$ of the quadrangle $ABCD$ intersect at point $O$ . We know that diagonal $BD$ is perpendicular to the side $AD$ , $\angle BAD = \angle BCD = 60^\circ$ , $\angle ADC = 135^\circ$ . Find the ratio $DO:OB$ .

Fig. 1
Answer: 1:2.

Solution

Under the problem statement we can easily find the following angles (fig.1): $\angle ABD = 30^\circ$ , $\angle BDC = 45^\circ$ , $\angle DBC = 75^\circ$ . Let's draw rays $ADE$ and $ABF$ . Then $\angle EDC = 45^\circ$ , $\angle FBC = 75^\circ$ . Therefore $BC$ is a bisector of $\angle DBF$ and $DC$ is a bisector of $\angle BDE$ , which implies that $AC$ is a bisector of $\angle BAD$ . The last statement is easily proved by the locus of the bisector. Thus $\angle BAO = 30^\circ$ and $\angle AOB$ is an isosceles triangle. Therefore $AO = BO$ and $DO = \frac{1}{2}AO$ as $\triangle ADO$ is right-angled triangle with an angle of $30^\circ$ . From this we find that $\frac{DO}{OB} = \frac{1}{2}$ .

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