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 $\cos A$ is equal to

• A. $\frac{ad+bc}{ad-bc}$
• B. $\frac{ad-bc}{ad+bc}$
• C. $\frac{ac+bd}{ac-bd}$
• D. $\frac{ac-bd}{ac+bd}$

Answer 1

The correct option is B $\frac{ad-bc}{ad+bc}$

Since circle is inscribed in the quadrilateral $ABCD$ then
$a+c=b+d$ ...........$\left(1\right)$
and quadrilateral is cyclic,then
$\angle A+\angle C=\pi$ or $\angle C=\pi -\angle A$ ..............$\left(2\right)$
Now, in $△ABD,$
$\cos A=\frac{a^2+d^2-{\left(BD\right)}^2}{2ad}$
or ${BD}^2=a^2+d^2-2ad\cos A$ ...................$\left(3\right)$
and in $△BCD,$
$\cos C=\frac{b^2+c^2-{\left(BD\right)}^2}{2bc}$
or ${BD}^2=b^2+c^2-2bc\cos C$
$=b^2+c^2-2bc\cos (\pi -\angle A)$ (from $\left(2\right)$)
${BD}^2=b^2+c^2+2bc\cos \angle A$ ................$\left(4\right)$
From eqns$\left(3\right)$ and $\left(4\right)$ we get
$2bc\cos A={BD}^2-b^2-c^2$
$2bc\cos A=a^2+d^2-2ad\cos A-b^2-c^2$
$\cos A=\frac{a^2+d^2-b^2-c^2}{2(bc+ad)}$ ..................$\left(5\right)$
Now, from eqn$\left(1\right),$
$a+c=b+d$
$\Rightarrow a-d=b-c$
Squaring both sides, we get
${(a-d)}^2={(b-c)}^2$
$a^2+d^2-2ad=b^2+c^2-2bc$
$\Rightarrow a^2+d^2-b^2-c^2=2ad-2bc$ ...............$\left(6\right)$
Substituting $\left(6\right)$ in $\left(5\right)$ we get
$\cos A=\frac{2ad-2bc}{2(ad+bc)}=\frac{ad-bc}{ad+bc}$