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 \(\frac{AC}{BD} = \frac{ps+qr}{pq+rs} \)

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

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

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