Question

In the figure, the tangent from A touches the circle at B. BC is diameter. The line AC cuts the circle at P.

The tangent at P cuts AB at M. Which of the following options are correct?

• A. $\angle PAM$ = $\angle PBM$
• B. $\angle PAM$ = $\angle APM$
• C. $\angle BPM$ = $\angle PBM$
• D. M is the midpoint of AB.

Answer 1

Answer: (B), (C), (D)

The correct options are
B

$\angle PAM$ = $\angle APM$

C

$\angle BPM$ = $\angle PBM$

D

M is the midpoint of AB.

Join PB. The figure looks like :

(i)
$\angle BPM$ = $\angle PBM$ = $\angle PCB$ = $x^{\circ }$ ...........(angles in the alternate segment are equal)

(ii)
$\therefore$ PM = MB ........(base angles of an isosceles $△$)

(iii)
$\angle CPB$ = ${90}^{\circ }$ ...........(angle in the semicircle)

(iv)
$\therefore$ $\angle APB$ = ${90}^{\circ }$ .........(linear pair)

(v)
$\angle APM$ = ${90}^{\circ }$ - $x^{\circ }$

(vi)
$\angle PMA$ = $2x^{\circ }$ ...........(exterior angle)

(vii)
$\angle PAM$ = ${90}^{\circ }$ - $x^{\circ }$ .........(Sum of angles of a $△$)

(viii) $\therefore$ $\angle APM$ = $\angle PAM$ = ${90}^{\circ }$ - $x^{\circ }$ ..........from (v) and (vii)

(ix) PM = AM .........(Base angles of an isosceles triangle)