Question

plz answer tomorrow is my paper. both parts plz

Answer 1

Answer:

Hi.  Hope this helps.

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Part A

∠TAB = ∠TCB    ( since TA is tangent )

=> ∠TAB + ∠BAD = ∠TCB + ∠CAD      ( AD bisects ∠CAB )

=> ∠TAD = ∠TDA       ( external angle of triangle )

=> ΔADT is isosceles

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Part B

AD ⊥ BC  and  AD² = BD . DC

=>  ∠ADC = ∠CDA (both right angles) and AD / BD = DC / DA

=> ΔADC is similar to ΔBDA

Then

∠BAC + ∠ACB + ∠CBA = 180°    ( sum of angles in a triangle )

=> ∠BAC + ∠ACD + ∠DBA = 180°    ( same angles )

=> ∠BAC + ∠BAD + ∠DAC = 180°    ( use the similar triangles above )

=> ∠BAC + ∠BAC = 180°

=> ∠BAC = 90°