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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

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