In the picture, a triangle is formed by two mutually perpendicular tangents to a circle and a third tangent.
Prove that the perimeter of the triangle is equal to the diameter of the circle.
Let us label the diagram and let the radius be 'r'.
To Prove : Perimeter of triangle PQR = diameter of circle
i.e. PQ + QR + PR = 2r
Construction: Join OA and OB
In Quadrilateral OAPB
∠OAB + ∠APB + ∠OBP + ∠AOB = 360°
∠OAB = 90° [OA ⊥ AP, as tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OBP = 90° [OB ⊥ BP, as tangent at any point on the circle is perpendicular to the radius through point of contact]
∠APB = 90° [AP ⊥ BP, Given]
⇒ 90 + 90 + 90 + ∠AOB = 360
⇒ ∠AOB = 90°
OA = OB [radii of same circle]
AP = BP [Tangents drawn from same point to a circle are equal]
And we know, if in a quadrilateral
i) All angles are 90° and
ii) Adjacent sides are equal
then the quadrilateral is square.
⇒ OAPB is a square
⇒ AP = BP = OA = OB = 'r' [radius] …
Also, ⇒ as tangents drawn from same points to a circle are equal
⇒ AQ = CQ [tangents from Q] …
⇒ BR = CR [tangents from R] …
Perimeter of triangle PQR
= PQ + QR + PR
= PQ + QC + CR + PR
= PQ + AQ + BR + PR [From 2 and 3]
= AP + BP
= r + r [From 1]
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