Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say we have two circles that intersect at points $X$ and $Y$. We also have a point $A$ on one of those circles, call it the "first circle", with a tangent line $T$ at the point $A$. Lines are extended from $A$ to $X$ and $Y$ until they intersect with the other circle, the "second circle" at points $B$ and $C$ respectively.

My end game is to prove that $BC$ and $T$ are parallel; here's what I have so far:

The tangent $T$ is perpendicular to the radius of the first circle. Extend this radius into a line, call it $K$. The chord $BC$ has a perpendicular bisector that passes through the centre of its circle. Extend this perpendicular bisector into a line, call it $L$.

It's clear that K and L are parallel, but I'm unsure of how to prove this.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Note that $BCYX$ is a cyclic quadrilateral and hence $$\underbrace{\angle BCY = \angle YXA}_{\text{Exterior angle of a cyclic quadrilateral equals the interior opposite angle}}\\ \underbrace{\angle YXA = \angle TAY}_{\text{External angle of a triangle is equal to the opposite interior angle of the triangle}}$$ Hence, $$\angle BCY = \angle TAY.$$ This means the alternate angle are equal and hence the lines are parallel.

share|improve this answer
    
That's a much easier way of going about this question. Thanks. –  Charlie Jun 5 '12 at 4:24

$$\angle TAY=\angle YXA=\angle BCY$$ Of course, the last equality gets you done as you've proved that alternate interior angles wrt lines $\,TA\,,\,BC\,$ are equal.

Now, how's the last equality proved? Look at the complementary angle of $\,\angle YXA\,\,,\,i.e.\,\,\angle BXY$, which is the opposite angle of a circumscribed quadrilateral in the leftmost circle...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.