# Chord-Tangent Theorem

• Dec 20th 2007, 08:30 PM
Truthbetold
Chord-Tangent Theorem
"We can prove that the measure of the angle formed by a chord and a tangent segment that intersect at the tangen'ts point of contact is equal to one half the measure of the intercepted arc."

Is this another of saying, with the proper geometric lines, intercepted arcs and inscribed angles work? That an inscribed angle is 1/2 of the intercepted arc.

Thanks!
• Dec 20th 2007, 10:07 PM
Whoever is giving you these should really have drawn you a picture.

C is the centre of the circle and A = 1/2 b

Attachment 4737
• Dec 21st 2007, 01:43 PM
Truthbetold
Quote:

Whoever is giving you these should really have drawn you a picture.

C is the centre of the circle and A = 1/2 b

Attachment 4737

I assume you mean b= 1/2a
or 2b=a.
Or not?
Thanks!
• Dec 22nd 2007, 12:27 AM
Quote:

I assume you mean b= 1/2a
or 2b=a.
Or not?
Thanks!
I stand by what I said originally. Looking at the picture, b is clearly bigger than A.
• Dec 27th 2007, 02:11 AM
mr fantastic
Quote:

I stand by what I said originally. Looking at the picture, b is clearly bigger than A.

badgerigar is correct - on two counts:

1. S/he is standing by something that is indeed correct (see below).
2. His/her remark "Whoever is giving you these should really have drawn you a picture" is spot on.

An equivalent statement of this theorem (as I understand it from the posts) is:

The angle between a tangent and a chord drawn from the point of contact is equal to any angle in the alternate segment.

The equivalence comes from a theorem that you (should) know: The angle subtended at the circumference is half the angle at the centre subtended by the same arc.
• Jan 2nd 2008, 12:40 PM
Henderson
It might be nice to have a quick explanation of why this works...

The triangle in your picture is clearly isosceles, since two of its sides are radii in the same circle (the third side being the chord you've drawn). With it being isosceles, the two non-b angles must be congruent, and will both have a measure of $\frac{180-b}{2}$, or $90 - \frac{b}{2}$.

Down at the point of tangency, we have a right angle, since tangents are perpendicular to the radius they touch. That right angle is composed of angle a, and one of our $(90 - \frac{b}{2})$ angles. So:

$a = 90 - (90 - \frac{b}{2})$

$a = 90 - 90 + \frac{b}{2}$

$a = \frac{b}{2}$

$QED$.