From a point P, two tangents PA and PB are drawn to a circle C(O,r). If OP=2r, show that △APB is an equilateral triangle.
Answer:
- Let OP meet the circle at Q. Join OA and AQ.
- We know that the radius through the point of contact is perpendicular to the tangent. So, OA⊥AP⟹∠OAP=90∘…(i)
- The circle is represented as C(O,r), this means that O is the center of the circle and r is its radius.
⟹OQ=OA=r…(ii)
Also, we see that OP=OQ+QP.
Substituting the value of OP and OQ in the above equation, we have OP=OQ+QP⟹2r=r+QP⟹QP=2r−r=r⟹ Q is the mid-point of OP. [As QP=OQ=r] - As, Q is the mid-point of OP,AQ is the median from the vertex A to the hypotenuse OP of the right-angled triangle AOQ.
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=OQ=QP=r⟹OA=OQ=QA=r[Using eq (ii)]⟹△AOQ is an equilateral triangle. ⟹∠AOQ=60∘ [Each angle of an equilateral triangle is 60∘]⟹∠AOP=60∘ [As ∠AOQ and ∠AOP is the same angle.] …(iii) - We know that the sum of angles of a triangle is 180∘.
For △AOP, ∠AOP+∠OAP+∠APO=180∘⟹60∘+90∘+∠APO=180∘ [Using eq (iii) and eq (i)]⟹∠APO=180∘−60∘−90∘=30∘ Also, two tangents from an external point are equally inclined to the line segment joining the center to that point.
So, ∠APB=2∠APO=2×30∘=60∘…(iv) - The lengths of the tangents drawn from an external point to a circle are equal.
So, PA=PB⟹∠PAB=∠PBA [Angles opposite to equal sides are equal.] …(v) - Consider △PAB ∠PAB+∠PBA+∠APB=180∘[Sum of angles of a triangle.]⟹∠PAB+∠PBA+60∘=180∘[Using eq (iv)]⟹2∠PAB=120∘[Using eq (v)]⟹∠PAB=60∘ Similarly, ∠PBA=60∘.
- As all the angles of the △PAB measure 60∘, it is an equilateral triangle.