From a point PPP, two tangents PAPAPA and PBPBPB are drawn to a circle C(O,r)C(O,r)C(O,r). If OP=2rOP=2rOP=2r, show that △APB△APB△APB is an equilateral triangle.
Answer:
- Let OPOPOP meet the circle at QQQ. Join OAOAOA and AQAQAQ.
- We know that the radius through the point of contact is perpendicular to the tangent. [Math Processing Error]
- The circle is represented as C(O,r)C(O,r), this means that OO is the center of the circle and rr is its radius.
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Also, we see that OP=OQ+QP.OP=OQ+QP.
Substituting the value of OPOP and OQOQ in the above equation, we have [Math Processing Error] - As, QQ is the mid-point of OP,AQOP,AQ is the median from the vertex AA to the hypotenuse OPOP of the right-angled triangle AOQAOQ.
We know that the median on the hypotenuse of a right- angled triangle is half of its hypotenuse.
Thus, QA=12OP=12(2r)=r.QA=12OP=12(2r)=r. [Math Processing Error] - We know that the sum of angles of a triangle is 180∘.180∘.
For △AOP△AOP, [Math Processing Error] Also, two tangents from an external point are equally inclined to the line segment joining the center to that point.
So, [Math Processing Error] - The lengths of the tangents drawn from an external point to a circle are equal.
So, [Math Processing Error] - Consider △PAB△PAB [Math Processing Error] Similarly, ∠PBA=60∘.∠PBA=60∘.
- As all the angles of the △PAB△PAB measure 60∘60∘, it is an equilateralequilateral triangle.