Question

In the given figure, $PM$ is a tangent to the circle and $PA=AM$, then,
(A) $△PMB$ is isosceles
(B) $PA\times PB={MB}^2$

• A. $A$ is true, but $B$ is false
• B. $B$ is true, but $A$ is false
• C. Both $A$ and $B$ are false
• D. Both $A$ and $B$ are true

Answer 1

The correct option is D Both $A$ and $B$ are true

$\left(A\right)$

In $△PAM$

$\Rightarrow$ $\angle APM=\angle AMP$ ----- ( 1 ) [ Base angles of equal sides are also equal ]

$\Rightarrow$ $PA=AM$ [ Given ]

By alternate segment property of tangent

$\Rightarrow$ $\angle ABM=\angle AMP$ ----- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $\angle APM=\angle ABM$

$\therefore$ $PM=MB$

Hence, $△AMP$ is an isosceles triangle.

$\left(B\right)$

By rectangle property of tangent and chord

$\Rightarrow$ $P{M}^2=PA\times PB$

$\Rightarrow$ $M{B}^2=PA\times PB$ [ Since, $PM=MB$ ]