Question

From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that Δ APB is equilateral.

Answer 1

Let us first put the given data in the form of a diagram.

Consider and. We have,

PO is the common side for both the triangles.

PA = PB(Tangents drawn from an external point will be equal in length)

OB = OA(Radii of the same circle)

Therefore, by SSS postulate of congruency, we have

ΔPOA ΔPOB

Hence,

…… (1)

Now let us consider and . We have,

PL is the common side for both the triangles.

(From equation (1))

PA = PB (Tangents drawn from an external point will be equal in length)

From SAS postulate of congruent triangles,

Therefore,

PL = LB …… (2)

Since AB is a straight line,

Let us now take up ΔOPB. We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,

By Pythagoras theorem we have,

It is given that

OP = diameter of the circle

Therefore,

OP = 2OB

Hence,

Consider . We have,

But we have found that,

Also from the figure, we can say

PL = PO − OL

Therefore,

…… (3)

Also, from , we have

…… (4)

Since Left Hand Sides of equation (3) and equation (4) are same, we can equate the Right Hand Sides of the two equations. Thus we have,

=

We know from the given data, that OP = 2.OB. Let us substitute 2OB in place of PO in the above equation. We get,

Substituting the value of OL and also PO in equation (3), we get,

Also from the figure we get,

AB = PL + LB

From equation (2), we know that PL = LB. Therefore,

AB = 2.LB

We have also found that. We know that tangents drawn from an external point will be equal in length. Therefore, we have

PA = PB

Hence,

PA =

Now, consider . We have,

PA =

PB =

AB =

Since all the sides of the triangle are of equal length, is an equilateral triangle. Thus we have proved.