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