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

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## 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

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

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## 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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