Proof: If two secants...intersect in the exterio of a circle

[josh90]

Write a two-column proof of Theorem 10.14: If two secants, a secant and a tangent, or two tangents interesect in the exterior of a circle, the measure of the angel formed is one-half the positive difference of the measures of the intercepted arcs.

Given: lines HI and HJ are tangents to circle O.

Prove: m. angle arc IHJ= one-half(m. arc IXJ-m. arc IJ

If someone could explain to me some steps in doing this problem I would really appreciate it. Thanks.

 

[stapel]

So we're looking at the same picture, draw the circle with the tangents leading off to their meeting point, H, to the right. Put I near the top of the circle, and J near the bottom. Put X somewhere off to the left.

Draw the segment HO. Recall that tangents, by definition, meet at right angles, and draw the right-angle symbols for angle OIH and OJH.

You now have two right triangles. Since these triangles have the same hypotenuse (namely OH) and same-length legs (namely OI and OJ), then, by the Pythagorean Theorem, the tangent segments IH and JH must have the same lengths. Thus, by SSS, you know that these triangles are congruent. Specifically, the angle at H is bisected by the line OH.

Label the angle at H as "theta". Then m(IHO) = (theta)/2 = m(JHO).

Label the remaining angle of each triangle as "(alpha)/2", where (alpha) is the measure of the angle subtending arc IJ (the arc on H's side of the circle). Label the angle for the other arc as "beta".

Since the triangles are right, what can you say about (alpha)/2 + (theta)/2?

Since they are inside a circle, what can you say about (alpha) + (beta)?

See where that takes you....

Eliz.