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# Ratio of diagonals of cyclic quadrilateral (TOMATO 110)

Let ABCD be a cyclic quadrilateral with lengths of sides AB = p , BC = q , CD = r, and DA = s . Show that $$\frac{AC}{BD} = \frac{ps+qr}{pq+rs}$$

Discussion:

Since ABCD is cyclic $$\Delta ABE$$ is similar to $$\Delta CDE$$ (since $$\angle ABE = \angle DCE$$ as they are subtended by the same arc AD, and $$\angle AEB = \angle CED$$ as vertically opposite angles are equal)

Hence their corresponding sides are proportional.

$$\frac{p}{r} = \frac{BE}{CE} \implies BE = CE \times \frac{p}{r}$$
$$\frac{p}{r} = \frac{AE}{DE} \implies AE = DE \times \frac{p}{r}$$

Similarly $$\Delta AED$$ is similar to $$\Delta BEC$$

$$\frac{s}{q} = \frac{DE}{CE} \implies DE = CE \times \frac{s}{q}$$
$$\frac{q}{s} = \frac{CE}{DE} \implies CE = DE \times \frac{q}{s}$$

Hence
$$BE + DE = BD = CE \times { \frac{p}{r} + \frac{s}{q} } = CE \times \frac{pq+rs}{qr}$$
$$AE + CE = AC = DE \times { \frac{p}{r} + \frac{q}{s} } = DE \times \frac{ps+qr}{sr}$$

Hence $$\displaystyle { \frac{AC}{BD} = \frac {ps + qr}{pq + rs} \times \frac {qr}{sr} \times \frac {DE}{CE} }$$

Finally we note that since $$\Delta AED$$ is similar to $$\Delta BEC$$. $$\displaystyle {\frac{q}{s} = \frac {CE}{DE} \implies \frac{q}{s} \times \frac {DE}{CE} = 1 }$$

This proves that

$$\displaystyle { \frac{AC}{BD} = \frac {ps + qr}{pq + rs}}$$

April 20, 2015