Triangle geometry, bisecting chords

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Source: BWM 2004, 2nd round, problem 3

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#1 h Sep 1, 2004, 10:10 PM • 2 Y

Y by Adventure10, Mango247

Given two circles $k_1$ and $k_2$ which intersect at two different points $A$ and $B$ . The tangent to the circle $k_2$ at the point $A$ meets the circle $k_1$ again at the point $C_1$ . The tangent to the circle $k_1$ at the point $A$ meets the circle $k_2$ again at the point $C_2$ . Finally, let the line $C_1C_2$ meet the circle $k_1$ in a point $D$ different from $C_1$ and $B$ .

Prove that the line $BD$ bisects the chord $AC_2$ .

This post has been edited 2 times. Last edited by darij grinberg, Oct 13, 2005, 11:16 AM

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#2 h Sep 1, 2004, 10:58 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the point of intersection of the lines $BD$ and $AC_2$ . We have to show that this point $M$ is the midpoint of the segment $AC_2$ .
Now, $<C_1AB=<BC_2A=<BDC_2, <AC_1B=<BAC_2$ , also: $MC_2B$ and $MDC_2$ are similar triangles: $MB.MD=MC_2^2$
As $AM$ is tangent to $k_1$ : $AM^2=MB.MD$
Hence: $AM=MC_2$ . QED.

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#3 h Sep 1, 2004, 11:09 PM • 2 Y

Y by Adventure10, Mango247

This is a very nice proof.

Here is my solution (I know that there are many different solutions, I've myself found three or four, but this is the one I have submitted):

Let N and M be the midpoints of the chords $AC_1$ and $AC_2$ , respectively. We will first show that the points A, B, N, M lie on one circle.

We will work with directed angles modulo 180°. Since the line $AC_2$ is the tangent to the circle $k_1$ at the point A, the tangent-chordal angle theorem yields $\measuredangle \left( AB;\;AC_{2}\right) =\measuredangle BC_{1}A$ . In other words, $\measuredangle BAC_{2}=\measuredangle BC_{1}A$ . Similarly, $\measuredangle BAC_{1}=\measuredangle BC_{2}A$ , or, in other words, $\measuredangle BC_{2}A=\measuredangle BAC_{1}$ . Together with $\measuredangle BAC_{2}=\measuredangle BC_{1}A$ , this yields that the triangles $BAC_2$ and $BC_1A$ are directly similar. Now, corresponding points in directly similar triangles form equal angles. The points M and N are corresponding points in our two directly similar triangles $BAC_2$ and $BC_1A$ , since M is the midpoint of the side $AC_2$ of the former triangle, while N is the midpoint of the side $C_1A$ of the latter one. Hence, $\measuredangle AMB=\measuredangle C_{1}NB$ . But $\measuredangle C_{1}NB=\measuredangle ANB$ . Thus, $\measuredangle AMB=\measuredangle ANB$ , and it follows that the points A, B, N, M lie on one circle. This, in turn, yields $\measuredangle ABM=\measuredangle ANM$ .

Now, since the points N and M are the midpoints of the sides $AC_1$ and $AC_2$ of triangle $AC_1C_2$ , we have $NM\parallel C_{1}C_{2}$ . Hence, $\measuredangle ANM=\measuredangle AC_{1}C_{2}$ . Together with $\measuredangle ABM=\measuredangle ANM$ , this gives $\measuredangle ABM=\measuredangle AC_{1}C_{2}$ .

Now, since the points A, B, D, $C_1$ all lie on the circle $k_1$ , we have $\measuredangle ABD=\measuredangle AC_{1}D$ , so $\measuredangle ABD=\measuredangle AC_{1}D=\measuredangle AC_{1}C_{2}=\measuredangle ABM$ . And this entails that the points B, D and M lie on one line. In other words, the line BD passes through the midpoint M of the chord $AC_2$ , i. e. it bisects this chord. Qed..

By the way, this problem has already been posted at MathLinks.

Darij

This post has been edited 4 times. Last edited by darij grinberg, Oct 13, 2005, 10:41 AM

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#4 h Sep 1, 2004, 11:45 PM • 2 Y

Y by Adventure10, Mango247

Let $d$ be the line through $D$ and parallel to $AC_2$ . We have to show that $DC_1,d,DA,DB$ form a harmonic quadruple, and after working on this a bit we reduce it to $\frac{BC_1}{BA}=\frac{AC_2}{AC_1}$ , which is easy to prove (it's in fact the ratio of the radii of the two circles).

Well, it's not that short, but this is the main idea

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