Question

AB is a diameter of a circle and c is a point on circumference.AD is perpendicular on the tangent drawn at c.prove that ac bisects angleBAD.

Answer 1

Refer the attachment for figure.



In the figure

AD is perpendicular to tangent drawn through C (given)

Draw a line from the centre O to the point of tangency, C

So, OC would be perpendicular to the tangent.
(line drawn from the centre to the tangent is perpendicular to it)

Now consider the lines AD and OC.

angle ADC = 90°
angle OCD = 90°

Angle ADC + Angle OCD = 90 + 90 = 180°

Angle ADC and angle OCD lies to the interior of AD and OC, and their sum is 90°

=> AD || OC

(if sum of co interior angles is 180°, then the lines are parallel to each other)

Now since AD || OC,

=> angle OCA = angle DAC .......(1)
(alternate angles)

Also, in ∆OAC,

OA = OC

=> angle OCA = angle OAC ......(2)

(In isoceles triangle, the angles opposite to equal sides are also equal)

But from (1) we know that

angle OCA = angle DAC

and from (2) we know that

angle OCA = angle OAC

equating the two we get

=> angle OAC =angle DAC

since these two angles are equal, this means that they are bisected by AC

=> AC is the angles bisector of angle OAD

now angle OAD coincides with angle BAD

=> AC is the angle bisector of angle BAD

Hence Proved