Track / Stage 4 / 302 of 340 #562 of 1964

Problem 562

AMC 12 late, AIME early

Geometry Difficulty 4.9 Multiple choice

Let $O$ be the intersection of diagonals $A C$ and $B D$ in quadrilateral $A B C D$ . If $\angle B A D + \angle A C B = 180^{\circ}$ , and $B C = 3, A D = 4, A C = 5, A B = 6$ , then $\frac{D O}{O B}=(\quad)$ .

Pick one

Official solution

4. A.

As shown in Figure 4, draw $B E / / A D$ , intersecting the extension of $A C$ at point $E$ .
$\begin{array}{l} \text { Then } \angle A B E=180^{\circ}-\angle B A D=\angle A C B \\ \Rightarrow \triangle A B C \backsim \triangle A E B \Rightarrow \frac{A C}{A B}=\frac{B C}{E B} \\ \Rightarrow E B=\frac{A B \cdot B C}{A C}=\frac{18}{5} . \end{array}$

Furthermore, since $B E / / A D \Rightarrow \frac{D O}{O B}=\frac{A D}{B E}=\frac{4}{\frac{18}{5}}=\frac{10}{9}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.