Question

show that the length of tangents drawn from an external point to a circle are equal

Answer 1

Hey there!

» 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: Join OA, OB, and OP.

» Proof: Since,
Tangent at any point of a circle is perpendicular to the radius through the point of contact.

OA ⊥ AP and OB ⊥ PB -------(1)

In Triangle, OPA and OPB:

OAP = OBP ----Using (1)

OA = OB (Radii of the same circle)

OP = OP (Common side)

Therefore, OPA ≅ OPB (RHS congruency criterion)

So, PA = PB (Corresponding parts of congruent triangles are equal)

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

HOPE IT HELPED ^_^

Answer 2

Given: AP an AQ are tangents from point A
To prove: AP=AQ
Construction: Join OP, OQ and OA
Proof:
In ΔOPA & ΔOQA,
OP⊥AP and OQ⊥AQ.( Tangent at any point of the circle is perpendicular to the radius through the point of contact.)
⇒∠OPA=∠OQA=90°······(1)
∴OP=OQ( radii of the circle)
  ∠OPA=∠OQA······(From 1)
  OA=OA(common)
By RHS congruence rule,
ΔOPA is congruent to ΔOQA
AP=AQ(by CPCT)
HENCE PROVED