Question

In a figure O is the centre of a circle. PA and PB are tangents forom an external point P touching the circle at points A and B. If OP is equal to the diameter then prove that ABP is an equilateral triangle.

Answer:

Given:
OP = length of diameter of the cirlce, O is the center and PA and PB are the two tangents to the circle.
Proof:
OA and OB are the radii of the circle.
OP = length of diameter = 2(length of radius) = 2(OA) = 2(OB) ...(1)
angle PAO = angle PBO = 90 degrees (Angles between the radii and tangents)
PO is common and OA = OB (Radii of circle)
Therefore, triangle PAO is congruent to triangle PBO.
Since angle PAO is 90 degrees, triangle PAO is a right triangle. Therefore,
sinθ = OA/OP
From (1), we get:
sinθ = OA/2(OA)
sinθ = 1/2
Therefore, θ = 30 degrees and so angle OPB = 30 degrees (CPCTC). Therefore, angle APB = 60 degrees.
PA = PB (CPCTC). Therefore, angle PAB = angle PBA.
Angle PAB + angle PBA + angle APB = 180 (angle sum property of a triangle)
2(angle PAB) = 2(angle PBA) = 180 - 60
2(angle PAB) = 2(angle PBA) = 120 which gives us: angle PAB = angle PBA = 60 degrees.
As all the angles of the triangle measure 60 degrees, triangle APB is an equilateral triangle.
Hence, proved.


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