Question

1.AD bisects angle A of triangle ABC, where D lies on BC and angle C is greater than angle B. Prove that angle ADB is greater than angle ADC.

2.In a triangle ABC, right-angled at B, BD is drawn perpendicular to AC.Prove that:
i) angle ABD = angle C ii)angle CBD = angle A

Answer 1

angle DAB = angle DAC as AD is the bisector of angle A.

     angle ADB = exterior angle to triangle ADC at D.
                     =  angle DAC + angle C
   angle ADC = exterior angle to triangle ADB  at D
                 =  angle DAB + angle B
                 = angle DAC + angle B              
  if angle C is greater, then  angle ADB is greater.
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2.  angle ABD + angle DBC = 90
     angle ABD = 90 - angle DBC
                     = 90 - (90 - angle C),  = angle C
  as in right angle triangle DBC, angle BDC = 90, so angle DBC = 90 - angle C.
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   similar to the above proof.

in right angle triangle ABD,  angle D = 90.
  so angle ABD + angle A = 90
     so angle ABD = 90 - angle A.

Now angle CBD =  90 - angle ABD = 90 - (90 - angle A)
                       = angle A.