#### Question

In the given figure PQ is a tangent to the circle at A, AB and AD are bisectors of `angleCAQ` and `angle PAC`. if `angleBAQ = 30^@. prove that:

1) BD is a diameter of the circle

2) ABC is an isosceles triangle

#### Solution

1) `angleBAQ = 30^@`

SinceAB is the bisector of `angleCAQ`

`=> angleCAB = angleBAQ = 30^@`

AD is the bisector of `angle CAP` and P-A-Q,

`angle DAP + angle CAD + angle CAQ = 180^@`

`=> angleCAD + anglle CAD + 60^@ = 180 ^@`

`=> angle CAD = 60^@`

So `angle CAD + angle CAB = 60^@ + 30^@ = 90^@`

Since angle in a semi-circle = 90°

⇒ Angle made by diameter to any point on the circle is 90°

So, BD is the diameter of the circle.

2) SinceBD is the diameter of the circle, so it will pass through the centre.

By Alternate segment theorem

`angle ABD = angle DAC = 60^@`

So, in `angle BMA`,

`angle AMB = 90^@` .........(UseAngleSumProperty)

We know that perpendicular drawn from the centre to a chord of a circle bisects the chord.

`=> angle BMA = angle BMC = 90^@`

In `triangleBMA` and `triangleBMC`

`angleBMA = angleBMC = 90^@`

BM = BM (common side)

AM = CM(perpendicular drawn from the centre to a chord of a circle bisects the chord.)

⇒ ΔBMA ≅ ΔBMC

⇒ AB = BC (SAS congruence criterion)

⇒ ΔABC is an isosceles triangle.