TA is a tangent to the circle from a point T and TBC is a secant to the circle. If AD is the bisector of∠CAB, prove thatΔADT is isosceles.

 

 

 

 

 

Given : TA is tangent to the circle, TBC is the secant AD is the bisector of ∠CAB

To prove : ΔADT is an isosceles triangle

Proof : In order to prove this, we will use alternate segment theorem

⇒ ∠TAB = ∠BCA  ...(1)

AD is bisector of ∠BAC ⇒ ∠BAD = ∠DAC  ...(2)

Now, ∠TAD = ∠TAB + ∠BAD

 = ∠BCA + ∠DAC  (using (1) and (2))

 = ∠DCA + ∠DAC

 = 180° – ∠CDA  [ in ΔCAD, ∠CAD + ∠DCA + ∠CDA = 180°]

⇒ ∠TAD = ∠TDA  [As ∠CDA + ∠TDA = 180° (linear pair)]

⇒ TD = TA  (sides opposite to equal angles are equal)

Hence, ΔADT is an isosceles Δ