Example - 7

Let \(ABCD\) be a convex cyclic quadrilateral with sides \(a, b, c, d\). If the semi-perimeter s of the quadrilateral is defined as \(\begin{align}s = \frac{{a + b + c + d}}{2}\end{align}\), show that the area of the quadrilateral is 

\[\Delta  = \sqrt {(s - a)(s - b)(s - c)(s - d)} \]

Solution: Recall that a cyclic quadrilateral is one that can be inscribed in a circle, or in other words, whose vertices are concyclic.

Using the cosine rule,

\[{l^2} = {a^2} + {b^2} - 2ab\cos B = {c^2} + {d^2} - 2cd\cos D \qquad \qquad \qquad ...(1)\]

Since \(ABCD\) is cyclic, \(\begin{align}\angle B + \angle D = \pi \end{align}\), so that \(\begin{align}\cos D =  - \cos B\end{align}\).

Using this in (1), we have

\[\begin{align}\cos B = \frac{{{a^2} + {b^2} - {c^2} - {d^2}}}{{2(ab + cd)}} \qquad \qquad \qquad  ...(2)\end{align}\]

The area of \(ABCD\) can be expressed as the sum of areas of \(\begin{align}\Delta ABC\;{\text{and}}\;\Delta ACD\end{align}\). Thus,

\[\begin{align}\Delta  = \frac{1}{2}ab\sin B + \frac{1}{2}\;cd\;\sin D\end{align}\]

Since \(\begin{align}\angle B + \angle D = \pi \end{align}\), we have

\[\begin{align}\Delta  = \frac{1}{2}(ab + cd)\sin B\end{align}\]

From (2),

\[\begin{align}   {\sin ^2}B & = 1 - {\cos ^2}B = (1 - \cos B)(1 + \cos B)\\   & = \frac{{\left[ {{{(a + b)}^2} - {{(c - d)}^2}} \right]\;\left[ {{{(c + d)}^2} - {{(a - b)}^2}} \right]}}{{4{{(ab + cd)}^2}}}\\   & = \frac{{(a + b + c - d)(a + b - c + d)(a - b + c + d)( - a + b + c + d)}}{{4{{(ab + cd)}^2}}}\\   & = \frac{{4(s - a)(s - b)(s - c)(s - d)}}{{{{(ab + cd)}^2}}}\\    \Rightarrow  \qquad \Delta & = \sqrt {(s - a)(s - b)(s - c)(s - d)}    \end{align}\]

Example - 8

\(AC\) and \(BD\) be \(\begin{align}\theta \end{align}\). Let \(ABCD\) be a quadrilateral, and let the angles between the diagonals.

(a) Show that the area of \(ABCD\) can be written as \(\begin{align}\Delta  = \frac{1}{2}\;AC \cdot BD \cdot \sin \theta \end{align}\)

(b) If \(ABCD\) is a convex cyclic quadrilateral, show that 

\[\begin{align}AC \cdot BD = AB \cdot CD + AD \cdot BC\end{align}\]

Solution:(a) Note that

\[{\rm{area}}\;(\Delta ACD) = \frac{1}{2} \cdot AC \cdot ED\sin \theta \]

and

\[{\rm{area}}\;(\Delta ABC) = \frac{1}{2} \cdot AC \cdot EB\sin \;(\pi  - \theta ) = \frac{1}{2}.AC \cdot EB\sin \theta \]

so that

\[\begin{align}\Delta  = \frac{1}{2} \cdot AC \cdot (ED + EB) \cdot \sin \theta  = \frac{1}{2}AC \cdot BD \cdot \sin \theta \end{align}\]

(b) We undertake a construction by reflecting \(B\) along the perpendicular bisector of the diagonal \(AC.;\)

Note that

\[\begin{align}    \angle EAD & = \frac{1}{2}\;\angle (ED)\\    & = \frac{1}{2}\;\angle (EC + CD)\\    & = \frac{1}{2}\;\angle (AB + CD) &  & {\rm{How?}}\\    & = \;\angle ADB + \angle CAD\\    & = \angle AFB = \theta \;({\rm{say}})    \end{align}\]

Make sure you are clear about how this result was obtained.

Also note by symmetry that area (ABCD) = area (AECD).

\[\begin{align}  {\rm{area}}\;(ABCD) & = \frac{1}{2}AC \cdot BD \cdot \sin \theta \;\;(\text{by the result of part - (a))}\\  {\rm{area}}\;(AECD) & = {\rm{area}}\,(\Delta AED) + {\rm{area}}(\Delta ECD)\\  & = \frac{1}{2}AE \cdot AD \cdot \sin \theta + \frac{1}{2}EC \cdot CD \cdot \sin (\angle ECD)\\   {\rm{But}}\;\angle ECD = \pi - \theta \;{\text{due}}\;{\text{to}}&\;AECD\;{\text{being}}\;{\text{cyclic}}{\rm{.}}\;{\rm{Thus,}}\\  {\rm{area}}\;(AECD) & = \frac{1}{2}\sin \theta \;(AE \cdot AD + EC \cdot CD)\\   & = \frac{1}{2}\sin \theta (AD \cdot BC + AB \cdot CD)\;(\text{from remarks in}\;{\text{the}}\;{\text{figure}})  \end{align}\]

Comparing the two areas, the required equality is established.

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