2. If two segments from the from the same exterior point are tangent to a circle, then the two segments are congruent. Given: and are tangent to and and , respectively Prove:
Step-by-step answer: We know that both triangles SME and SLE share a hypotenuse, which is line SE. Since, we know that the definition of tangent to a circle mean that a line segment (in this case EM and EL) form a 90 degree angle with the radius, meaning that the base of the triangles are the radius. From the Pythagorean Theorem, sqrt(a^2 + b^2) = c, where c is the hypotenuse. Since, they share a hypotenuse, they have the same c. Now, let a be the base, which is r, the other side be b, and the hypotenuse be h, so substitute: Triangle SME: r^2 + b^2 = h^2 b = sqrt(r^2 + h^2) Triangle SLE: r^2 + b^2 = h^2 b = sqrt(r^2 + h^2) So in conclusion: Triangle SME: base = r (given) hypotenuse = h (given) other side = sqrt(r^2 + h^2) Triangle SLE: base = r (given) hypotenuse = h (given) other side = sqrt(r^2 + h^2) Therefore, by definition, triangles with same sides are congruent, so triangles SME and SLE are congruent since their sides are the same. The base, hypotenuse, and other sides are r, h, and sqrt(r^2 + h^2), respectively. |