Question

A quadrilateral $ABCD$ in which $AB=a,BC=b,CD=c$ and $DA=d$ is such that one circle can be inscribed in it and another circle circumscribed about it, then $\cos A=\frac{ad+bc}{ad-bc}$.

• A. True
• B. False

Answer 1

The correct option is B False

Since a circle can be inscribed in the quadrilateral, we have
$a+c=b+d$
and since the quadrilateral is cyclic,
$C=\pi -A$
$\cos A=\frac{a^2+d^2-{BD}^2}{2ad}$
$\Rightarrow 2ad\cos A=a^2+d^2-{BD}^2$
$\Rightarrow 2ad\cos A=a^2+d^2-[b^2+c^2-2bc\cos C]$
$=a^2+d^2-b^2-c^2+2bc\cos (\pi -A)$ $\because C=\pi -A$
$=a^2+d^2-b^2-c^2+2bc\cos A$
$\Rightarrow 2(ad+bc)\cos A=a^2+d^2-b^2-c^2$
$\cos A=\frac{a^2+d^2-b^2-c^2}{2(ad-bc)}$
Now, $a+c=b+d\Rightarrow a-d=b-c$
Now, $a^2+d^2-b^2-c^2$
$=[a^2+d^2]-[b^2+c^2]$
$=[{(a-d)}^2+2ad]-[{(b-c)}^2+2bc]$
$=[{(a-d)}^2+2ad-{(a-d)}^2-2bc]$ from above condition $a-d=b-c$
$=2ad-2bc$ on simplification
$=2(ad-bc)$
$\therefore \cos A=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}$
$=\frac{2(ad-bc)}{2(ad+bc)}$
$=\frac{(ad-bc)}{(ad+bc)}$