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}

Screen Shot 2015-04-21 at 1.34.06 AMDiscussion:

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}}