Question

Prove that the length of two triangles drawn from an external point to a circle are equal.​

Answer 1

Answer:

The length of two triangles drawn from an external point to a circle are equal.✔️

Step-by-step explanation:

Let two tangent AF and BF are drawn to circle of centre E as shown in figure.

Both the given tangents AF and BF touch to the circle at A and B respectively.

To Prove : length of AF = length of BF

Construction :- Draw a line segment ,from centre E to external point F.

Proof :

Now ∆AEF and ∆BEF

We know, tangent makes right angle with radius of circle.

Here, AE and BE are radii . So, ∠EAF = ∠EBF = 90°

Now, it is clear that both the triangles ∆AEF and BEF are right angled triangle.And a common hypotenuse EF of these [ as shown in figure ]

Now, come to the concept ,

∆AEF and ∆BEF

∠EAF = EBF = 90°

Common hypotenuse EF

And EA = EB [ EA and EB are radii]

So, [R - H - S] rule of similarity

∆AEF ~ ∆BEF [C. P. C. T]

Hence, EA/EB = AF/BF = EF/EF

AF/BF = 1

AF = BF [ hence proved]

Answer 2

$\huge\underline\mathbb {SOLUTION:-}$

Given:

A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.

To Prove:

• PA = PB

Construction:

1. Join OA, OB, and OP.

★Figure provided in the attachment.

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\therefore \mathsf {OA\:\perp PA\:and\:OB\:\perp PB..... (1)}$

$\mathsf {In\:\Delta OPA\:\Delta OPB\: :}$

$\mathsf {\angle OPA = \angle OBP\:(Using\:(1))}$

$\mathsf {OA = O B\:(Radii\:of\:the\:same\:circle)}$

$\mathsf {OP = OP\:(Common\:side)}$

$\therefore \mathsf {\Delta OPA \cong \Delta OPB\:(RHS\:congruency\:criterion)}$

$\therefore \mathsf {PA = PB}$

(Corresponding parts of congruent triangles are equal.)

Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.