Question

In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that :

(i) $\angle$ BAP = $\angle$ ADQ

(ii) $\angle$ AOB = 2$\angle$ ADQ

(iii) $\angle$ ADQ = $\angle$ ADB.

Answer 1

i) Since $PAT\parallel BC$

$\angle PAB=\angle ABC\left(alternateangles\right)$.....(i)

In cyclic quadrilateral ABCD ,

$Ext\angle ADQ=\angle ABC$........(ii)

from (i) and (ii) ,

$\angle PAB=\angle ADQ$

ii) Arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.

$\angle AOB=2\angle ADB$

$\to \angle AOB=2\angle PAB$ (angles in alternate angles)

$\to \angle AOB=2\angle ADQ$ (proved in (i) part)

(iii)$\angle BAP=\angle ADB$ (angles in alternate angles)

But,

$\angle BAP=\angle ADQ$

$\therefore \angle ADQ=\angle ADB$