Q1) ABC is an isoceles triangle in which AB=Ac circumscribed about a circle.show that BC is bisected at the point of contact.
Q2)The two tangents from an external point P to a circle with centre O are PA and PB. If angle APB=70degree, what is the value of angle AOB?
Q3) OP is equal to diameter of the circle.Prove that ABP is an equialteral triangle.
Q1.
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We know that the tangents drawn from an exterior point to a circle are equal in length.
∴ AP = AQ (Tangents from A) ..... (1)
BP = BR (Tangents from B) ..... (2)
CQ = CR (Tangents from C) ..... (3)
Now, the given triangle is isosceles, so given AB = AC
Subtract AP from both sides, we get
AB – AP = AC – AP
⇒ AB – AP = AC – AQ (Using (1))
BP = CQ
⇒ BR = CQ (Using (2))
⇒ BR = CR (Using (3))
So BR = CR, shows that BC is bisected at the point of contact.
Q2.
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Given:
∠ APB = 70° ..... (1)
and PA and PB be two tangents draw from an external point P to a circle with centre O.
OA ⊥ PA and OB ⊥ PB [ A tangent to a circle is perpendicular to the radius through the point of contact.]
Now, in right triangles
OAP and OBP, we have
OA = OB (Equal radius of the circle)
PA = PB (Tangents drawn from an external point are equal)
OP = OP (Common)
∴ ∆OAP ≅ ∆OBP (SSS)
⇒∠ AOP = ∠ BOP and ∠ OPA = ∠ OPB
∴ ∠ AOB = 2 ∠ AOP and ∠ APB = 2 ∠ OPA ..... (2)
In ∆ OPA, ∠ AOP + ∠ OAP + ∠OPA = 180°
⇒ ∠ AOP + ∠ OPA = 90° (as ∠ OAP = 90°)
∴ ∠ AOP = 90° – ∠ OPA
multiply by 2 on both sides
2∠ AOP = 180° 2∠ OPA
∠ AOB = 180° – ∠ APB (Using (2))
Substituting the values from (1), we get
∠ AOB = 180° – 70° = 110°
3.)
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