[1]If PA and PB are tangents from an outside point P such that PA = 10 cm and angle APB = 60 .find the length of the chord AB.

[2]prove that the tangent drawn at the ends of a chord of a circle make equal with the chord.

[3]prove that the line segment joining the points of contact of two parallel tangent to a circle is a diameter of the circle.

Dear Student!

Here is the answer to your query.

**(1) Given **: PA and PB are tangents of a circle, PA = 10 cm and ∠APB = 60°

Let O be the center of the given circle and C be the point of intersection of OP and AB

In ΔPAC and ΔPBC

PA = PB ( Tangents from an external point are equal)

∠APC = ∠BPC ( Tangents from an external point are equally inclined to the segment joining center to that point)

PC = PC ( Common)

Thus ΔPAC ΔPBC (By SAS congruency rule) ..........(1)

∴ AC = BC

Also ∠APB = ∠APC + ∠BPC

∠ACP + ∠BCP = 180°

Now in right triangle ACP

∴ AB = AC + BC = AC + AC ( AC = BC)

⇒ AB = (5 + 5) cm = 10 cm

**(2) **From (1) we have

∠PAC = ∠PBC

**(3)**

**Given :** Tangent *l *is parallel to tangent *m* end A and B are its points of intersection with circle of center O and radius *r*.

**To Prove : **AB is the diameter

Let* l *and *m* meet at some point P

Then in quadrilateral AOBP

∠AOB + ∠OAP + ∠OBP + ∠APB = 360° ......(2)

We know that tangents to a circle is perpendicular to the radius

⇒ OA ⊥ *l *and OB ⊥ *m*

⇒ ∠OAP = 90° and ∠OBP = 90° .........(3)

Since *l* || *m *

*∴ *∠APB = 0° ........(4)

From (2), (3) and (4)

∠AOP + 90° + 90° + 0° = 360°

⇒ ∠AOB = 360° – 180° = 180°

*∴ *AB = AO + OB = *r + r* = 2*r* = Diameter

Hence AB is the diameter of the circle.

Cheers!

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