Question 787110: I have an image of a circle (center "O") with a triangle APB intersecting it. Line segment AB runs through the center "O" (pretty much the diameter of the circle.) Line segment AP is tangent to the circle. In addition, there is line segment AC that bisects triangle APB, with points A and C being on the perimeter of the circle.
Given: line segment PA is tangent to the circle with center "O"
Prove: Triangle APB is congruent to triangle CPA
I think the way to go with this is to identify < ACB and < ACP as right angles, but I don't know the rest and I really have trouble identifying the reasoning for each step.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Your explanation falls apart when you said "(pretty much the diameter of the circle)". It either is or is not the diameter. If both A and B are on the circle and AB passes through O, then it is a diameter. If B is somewhere other than on the circle, then it is not a diameter. You will then need to specify just where point B lies. Then you said "...segment AC that bisects triangle APB, ..." which doesn't make any sense either because you cannot "bisect" a triangle. You can bisect any of the angles of a triangle, and I suspect you mean that AC bisects angle A -- presuming you are using the term "bisect" correctly. Send me an e-mail with a scan of your diagram.
I'll give you one hint to start: Since AB passes through O, AO must be a radius. Since AO is a radius it is guaranteed to be perpendicular to a tangent, namely AP, at A. Hence angle PAB is indeed a right angle.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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