# Proving the converse to the tangent secant theorem

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#1
Let M be a point not on the circle and let MC be a line intersecting the circle at C.
Let MA be a secant to the circle, intersecting the circle at B and A.
Let |MC|2 = |MA||MB|.
Prove that MC is tangent to the circle at C.

How do I do this? I know that if angle MCB is congruent to angle CAB then that shows that MC is tangent to the circle by another theorem, but I don't know how to show that congruence.
0
6 years ago
#2
(Original post by UpstairsMuffin)
Let M be a point not on the circle and let MC be a line intersecting the circle at C.
Let MA be a secant to the circle, intersecting the circle at B and A.
Let |MC|2 = |MA||MB|.
Prove that MC is tangent to the circle at C.

How do I do this? I know that if angle MCB is congruent to angle CAB then that shows that MC is tangent to the circle by another theorem, but I don't know how to show that congruence.
Let F be the centre of the circle, and construct MF. Let E be such that ME is tangent to the circle (put it on the other side for clarity). Apply the tangent secant theorem to ME and MBA. Now start recognizing congruences.
1
6 years ago
#3
(Original post by UpstairsMuffin)
Let M be a point not on the circle and let MC be a line intersecting the circle at C.
Let MA be a secant to the circle, intersecting the circle at B and A.
Let |MC|2 = |MA||MB|.
Prove that MC is tangent to the circle at C.

How do I do this? I know that if angle MCB is congruent to angle CAB then that shows that MC is tangent to the circle by another theorem, but I don't know how to show that congruence.
Well, we know the converse is true (alternate segment theorem does the job) ie, if MD is tangent, then MD^2=MAMB. Consider now the circle with radius MD center M. This intersects the circle at most 2 times, and by symmetry, it intersects the circle at D' such that MD' is also tangent. Hence, at any other point P on the circle, MP=/= MD unless P=D or D'
0
#4
I'd like to use the theorem that if angle MCB is congruent to angle CAB then MC is tangent to circle ABC, though.
Is there some way I could use similarity?
0
#5
Eh alright, I'll just use the proofwiki proof. Thanks.
0
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