Two tangents PQ and PR are drawn to a circle with center o from an external point P. prove that Angle QPR = 2 angle OQR

Refer the digram & explanation.

Step-by-step explanation:

GIVEN :-

TO PROVE :-

CONSTRUCTION :-

FACTS TO KNOW BEFORE SOLVING :-

PROCEDURE :-

∠QPR + ∠QOR = 180°

⇒ ∠QOR = 180° - ∠QPR

In ΔQOR , OQ = OR (∵ Radii of the circle are equal)

⇒ ΔQOR is an isosceles triangle (∵ OQ = OR i.e. two sides of the triangle are equal.)

⇒ ∠OQR = ∠ORQ (∵ In an isosceles triangle , the base angles are equal)

So,

∠QOR + ∠OQR + ∠ORQ = 180°

⇒ ∠QOR + 2∠OQR = 180°  (∵ ∠OQR = ∠ORQ)

⇒ 180° - ∠QPR = 180° - 2∠OQR  (∵ ∠QPR + ∠QOR = 180°)

⇒ -∠QPR = -2∠OQR  (∵ Cancelling 180° from both the sides.)

⇒ ∠QPR = 2∠OQR (Proved)

a

step-by-step explanation:

a

squares, rhombus, rectangles