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February 16, 2021

Test of Mathematics Solution Subjective 110 - Ratio of Diagonals of Cyclic Quadrilateral

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 110 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


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Problem:

:
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} $


Solution:

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

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

$ \frac{s}{q} = \frac{DE}{CE} \Rightarrow DE = CE \times \frac{s}{q} $
$ \frac{q}{s} = \frac{CE}{DE} \Rightarrow 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} \Rightarrow \frac{q}{s} \times \frac {DE}{CE} = 1 } $

This proves that

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

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