Question

From a point P , two tangents PA and PB are drawn to a circle with ​centre O. If OP is the diameter of the circle , show that ∆APB is equilateral.

Answer 1

Suppose OP meets the circle at Q. Join AQ.
OP=diameter of the circle
OQ+PQ=diameter
PQ=diameter-OQ
PQ=diameter-radius
PQ=2radius-radius=radius
Therefore,OQ=PQ=radius=OA....................... 1
We know that tangent is perpendicular to radius at point of contact,
therefore, angle OAP=90
Triangle OAP is a rigt angled triangle,
and OP is hypotenuse and also Q is the midpoint of OP,.
we known that Midpoint of hypotenuse of a right angled triangle is equidistant from the vertices.
therefore, AQ=OQ
from equation 1 we get,
AQ=OQ=OA
Therefore triangle OAQ is equilateral triangle.
therefore, angle AOQ=60
then angle APO=30(because it is half of that angle)
we know that angle APB=2angleAPO=60
we have,
PA=PB
Then, angle PAB=anglePBA as sides opposite to equal angles.
We have angle APB=60
By angle sum property of a triangle we get,
angle APB+PAB+PBA=180
anglePBA=anglePAB=60=angle APB
Hence triangle APB is an equilateral triangle.

Answer 2

∠OAP = 90°(PA and PB are the tangents to the circle.)

In ΔOPA,
sin ∠OPA = OA/OP  =  r/2r   [OP is the diameter = 2*radius]
sin ∠OPA = 1 /2 =  sin  30 ⁰

∠OPA = 30°

Similarly,

∠OPB = 30°.

∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB    (tangents from an external point to the circle)

⇒∠PAB = ∠PBA ............(1)   (angles opp.to equal sides are equal)

∠PAB + ∠PBA + ∠APB = 180°    [Angle sum property]

⇒∠PAB + ∠PAB = 180° – 60° = 120°  [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60°    .............(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

all angles are equal in an equilateral triangle.

ΔPAB is an equilateral triangle

hope it helps!!!!