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.