MTG 3212
Homework 6
February 18, 2009
5) a) “Useful Construction Theorem 11” says that if a triangle is inscribed in a circle with
one side as the diameter, then the opposite angle is 90◦ . Prove this. It’s really not too
difficult.
Ans: Let △ABC be the inscribed triangle with AB the diameter. Let O be the center. The
triangles △AOC and △COB are both isosceles triangles, which says the angle at C is the
sum of the angles at A and B. That forces the angle at C to be a right angle.
b) Now the other way: let △ABC be a right triangle with C = 90◦ and let O be the
midpoint of AB. Prove that CO ∼ AO ∼ BO (in other words, C is on the circle centered
= =
at O with AB as its diameter).
Ans: Add a point D to get yourself a rectangle. Then let O be the intersection of the
diagonals, and since this is a rectangle, O is equidistant from all four corners. This is what
you want to show.
6) Let C be a point on a circle whose center is O. Assume that line l passes through C, and
it is NOT perpendicular to OC. Show that the line l must intersect the circle at another
point. This indirectly proves “Useful Construction Theorem 12”: the line tangent to a circle
at a point C is perpendicular to the radius. Problem 5 would be really useful here.
Ans: If l is not perpendicular to OC, then one of the angles it makes with OC has a
measurement less than 90◦ , say α degrees. Extend OC to a diameter, and say the opposite
point is D. We can construct a line m at D such that it makes an angle whose measure
is 90◦ − α, and we can put this angle on the same side as the angle made with l and OC.
The lines l and m are not parallel, so they will intersect at a point A. By construction, the
triangle ACD will be a right triangle, and so we can use 5 b) to say that OC ∼ OD ∼ OA.
= =
This means A is on the circle, and so line l intersects the circle.