Question

TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Then prove that AP bisects $\angle TAB$.

Answer 1

Given - A circle with centre C. From a point T outside the circle, TA and TB are two tangent to the circle OT intersects the circle at P, AP and AB are joined.

To Prove AP is the bisector of $\angle TAB$

Construction - Join PB.

Proof - In $\Delta OATand\Delta OBT,$

AT = BT (Tangents from T to the circle)

OA=OB (Raddii of the same circle)

OT= OT (Common)

$\therefore \Delta OAT=\Delta OBT$ (SSS postulate)

$\angle ATO=\angle BTO$ (C.P.C.T.)

Or $\angle ATP=\angle BTP$

Now, in $\Delta APT$ and $\Delta BPT,$

AT = BT (Tangents from T)

PT = PT (Common)

$\angle ATP=\angle BTP$ (Proved)

$\therefore \Delta APT\cong \Delta BTP$ (S.A.S. postulate)

$\therefore \angle PAT=\angle PBT$ (C.P.C.T.)

and AP = BP (C.P.C.T)

$\therefore \angle PAB=\angle PBA$

(Angles opposite to equal sides)

But $\angle PAT=\angle PBA$ (Angles in alt. segment)

$\therefore \angle PAB=\angle PAT$

$\therefore$ AP is the bisector of $\angle TAB$.

Q.E.D