**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.

**Solution:**

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

$\triangle \mathrm{POA} \cong \triangle \mathrm{POB}$

Hence,

$\angle O P A=\angle O P B$............(1)

Now let us consider $\triangle P L A$ and $\triangle P L B$. 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)

$\angle P L A=\angle P L B$

Since AB is a straight line,

$\angle A L B=180^{\circ}$

$\angle P L A+\angle P L B=180^{\circ}$

$2 \angle P L A=180^{\circ}$

$\angle P L A=90^{\circ}$

$\angle P L B=90^{\circ}$

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,

$\angle O B P=90^{\circ}$

By Pythagoras theorem we have,

$P B^{2}=O P^{2}-O B^{2}$

By Pythagoras theorem we have,

$P B^{2}=O P^{2}-O B^{2}$

It is given that

*OP* = diameter of the circle

Therefore,

*OP = 2OB*

Hence,

$P B^{2}=(2 O B)^{2}-O B^{2}$

$P B^{2}=4 O B^{2}-O B^{2}$

$P B^{2}=3 O B^{2}$

$P B=\sqrt{3} O B$

Consider $\triangle P L B$. We have,

$L B^{2}=P B^{2}-P L^{2}$

But we have found that,

$P B=\sqrt{3} O B$

Also from the figure, we can say*PL = PO − OL *

Therefore,

$L B^{2}=3 O B^{2}-[P O-O L]^{2} \ldots \ldots(3)$

Also, from , we have

$L B^{2}=O B^{2}-O L^{2}$

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,

$O B^{2}-O L^{2}=3 O B^{2}-[P O-O L]^{2}$

$O B^{2}-O L^{2}=3 O B^{2}-\left[P O^{2}+O L^{2}-2 . P O . O L\right]$

$O B^{2}-O L^{2}=3 O B^{2}-P O^{2}-O L^{2}+2 . P O . O L$

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

$O B^{2}-O L^{2}=3 O B^{2}-(2 O B)^{2}-O L^{2}+2.2 . O B . O L$

$O B^{2}-O L^{2}=3 O B^{2}-4 O B^{2}-O L^{2}+4 . O B . O L$

$2 O B^{2}=4 . O B . O L$

$O L=\frac{O B}{2}$

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

$L B^{2}=3 O B^{2}-\left[2 O B-\left(\frac{O B}{2}\right)\right]^{2}$

$L B^{2}=3 O B^{2}-\left[4 O B^{2}+\left(\frac{O B^{2}}{4}\right)-2 . O B^{2}\right]$

$L B^{2}=3 O B^{2}-4 O B^{2}-\left(\frac{O B^{2}}{4}\right)+2 . O B^{2}$

$L B^{2}=\frac{3 O B^{2}}{4}$

$L B=\frac{\sqrt{3} O B}{2}$

Also from the figure we get,

*AB = PL + LB*

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

*AB = 2.LB*

$A B=2 \times \frac{\sqrt{3} O B}{2}$

$A B=\sqrt{3} O B$

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.

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