A cyclic quadrilateral has sides of lengths 2, 3, 4, and 6 units, respectively. In what ratio do the diagonals divide each other?
The vertices of the quadrilateral are $A, B, C, D$, and the intersection point of the diagonals is $M$. Angles marked the same on the diagram are equal because they are inscribed angles subtending the same arc. Therefore, the following two similarities hold:

$$
D M C \triangle \sim A M B \triangle \quad \text { and(1) } B M C \triangle \sim A M D \triangle .(2)
$$

In the first case, the similarity ratio is $\frac{6}{3}$, and in the second case, it is $\frac{4}{2}$, both equal to 2. Therefore, based on (1), with $C M=a$, we have $B M=2 a$, and with $D M=b$, we have $A M=2 b$. According to (2), $\frac{D M}{C M}=\frac{b}{a}=2$, so

$$
\frac{A M}{M C}=\frac{2 b}{a}=2 \cdot 2=4,(3) \text { and } \frac{D M}{M B}=\frac{b}{2 a}=\frac{1}{2} \cdot 2=1 .(4)
$$

(3) means that the diagonal $B D$ divides $A C$ in the ratio $4: 1$, and (4) means that $A C$ divides $B D$ in the ratio $1: 1$, or bisects it.

Garai Zsolt (Dombóvár, Illyés Gy. Gimn., 10. o.t.)

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