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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.
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