Question

Two circle intersect at $A$ and $B$. Quadrilaterals $PCBA$ and $ABDE$ are inscribed in these circles such that $PAE$ and $CBD$ are line segments. Also, $\angle$P = 95${}^o$ and $\angle$C = 40${}^o$. The value of $Z$ is:

• A. ${65}^o$
• B. ${105}^o$
• C. ${95}^o$
• D. ${85}^o$

Answer 1

Answer: (D)

${85}^o$

Given, $AB$ is the common chord of two intersecting circles.

$ABCP$ & $ABDE$ are two cyclic quadrilaterals inscribed in the circles in such a way that $\overline{PAE}$ & $\overline{CBD}$ are line segments,

i.e. $\overline{PAE}$ as well as $\overline{CBD}$ are straight lines.

Also, $\angle APC={95}^o$ & $\angle BCP={40}^o.$

Then, $\angle APC+\angle ABC={180}^o$.......(i) [since the sum of the opposite angles of a cyclic quadrilateral is ${180}^o$]

and $\angle ABC+\angle AED={180}^o$.......(ii) (linear pairs).

So, from (i) & (ii), we get,

$\angle APC=\angle ABD={ 95 }^{ o }.

Again $\angle ABD+\angle AED(=Z)={180}^o$...[since the sum of the opposite angles of a cyclic quadrilateral is ${180}^o$

$\therefore Z={180}^o-\angle ABD={180}^o-{95}^o={85}^o.$

Hence, option $D$ is correct.