Two tangents to a circle with center $O$ meet at $A$ and extent an angle of $50^\circ$. Another tangent to the circle intersects the two tangents at $B$ and $C$. If $\angle BOC = x^\circ$, find $x$.
+118673 If you take the extreme case where BC and BA are the same tangent.
then angle ABO is 90 degrees, angle BAO=25 degrees so angle x must be 180-25-90 = 65 degrees.
There is obviously some theory that Jugoslav knows that says this will always be the case.
I don't know how to prove the general case.
+397 Extend AB to B' and AC to C'.
Call angle CBO alpha, then angle OBB' = alpha, so angle ABC = 180 - 2(alpha).
Call angle BCO beta, then angle ACB = 180 - 2(beta).
From triangle ABC, (alpha + beta) = 115 deg, and then from triangle BOC, x = 65 deg.
+2489 Wonderful Tiggsy!
Except, without a bloody diagram, I’m lost to where B’ and C’ are supposed to be, relative to the circle. Are B’ and C’ positioned such that when they are joined by a vertical line, the line forms a tangent on the opposite side of the circle, mirroring the tangent of the line BC?
GA
+33661 I suspect Tiggsy meant the following for B' (with an analogous point for C'):
+118673 Thanks Tiggsy,
I still have a question,
How do you know that
angle CBO = angle OBB' ?
Here is the pic to go with Tiggsy's explanation.
It is interactive, you can grab the point T and move it around to see that the result says the same.
https://www.geogebra.org/classic/hqcjnaku
Here is the non interactive version:
+129852 .....
I still have a question,
How do you know that
angle CBO = angle OBB' ?....
Yeah.....I was wondering about that too....
+118673 I get it Chris, it is because BB' and BC are both tangents and BO goes through the centre of the circle, so
angle CBO and angle OBB' have to be congruent.
You could use the 3 equal sides test of congruency.
Plus if two tangents intersect then the distances from the cross point to the circle intersections will be congruent.
