Question

In Fig. TA is a tangent to a circle from the point T and TBC is a secant to the circle. If AD is the bisector of ∠BAC, prove that ΔADT is isosceles.

Answer 1

In order to prove that ΔADT is isosceles i.e., TA = TD, it is sufficient to show that ∠TAD =  ∠TDA.

Since ∠TAB and ∠BCA are angles in the alternate segments of chord AB.
∴ ∠TAB = ∠BCA       ...(i)

It is given that AD is the bisector of ∠BAC.
∠BAD = ∠CAD        ...(ii)

Now, ∠TAD = ∠TAB + ∠BAD 
⇒ ∠TAD = ∠BCA + ∠CAD      ....(Using (i) and (ii))
⇒ ∠TAD = ∠DCA + ∠CAD     ....( ∵∠BCA = ∠DCA)
⇒ ∠TAD = 180° - ∠CAD         ....( In ΔCAD, ∠CAD + ∠DCA +∠CDA = 180° ∴ ∠CAD + ∠BCA = 180° - ∠CAD)
⇒ ∠TAD = ∠TDA                  ....(∵∠CDA + ∠TDA = 180°)
= TD = TA
Hence, ΔADT is isosceles
Hence proved.