Proving Length of Chord in a Circle is Constant

Question

In a circle, a fixed chord, AB, is drawn. Point P is an arbitrary point on the major arc of AB and the chords AP and BP are drawn. The angle bisectors of $\angle PAB$ and $\angle PBA$ are drawn and extended so that they meet the circle at points C and D respectively. Prove that CD has a fixed length.

Lets call the point where the two angle bisectors meet as M. My first thought was that I had to prove that triangle AMB and triangle CDM were similar and then prove that the sides are at a fixed ratio to each no matter where P is. Proving that they are similar is straight forward, but I can't prove that the sides must be in equal ratios.

Then I tried to do the same thing for triangles CDP and ABM, but I also couldn't prove that they must in equal ratios.

Am I on the right track? Is using similar triangles the way to go for this question? Or I am totally on the wrong track and should be using other properties of the circle such as joining the chords to the radius and working from there?

Answer 1

Let $\angle PAB=2\alpha$, $\angle PBA=2\beta$, $\angle APB=\angle ACB=\pi-2(\alpha+\beta)=\gamma$:

Note that $\gamma=\mathbf{const}=\tfrac12\arccos\frac{2R^2-|AB|^2}{2R^2}$.

Then $\angle AMD=\alpha+\beta=\delta=\angle BMC$, hence $|AD|=|MD|$ $(\triangle AMD)$, $|MC|=|BC|$ $(\triangle BMC)$, and since $\angle AMB=\angle DBC$ it follows that $\triangle AMB\sim\triangle DMC$,

\begin{align} \frac{|CD|}{|AB|} &= \frac{|MD|}{|MA|} = \frac{|MC|}{|MB|} = \phi. \end{align}

From $\triangle DAM$, $|MD|\cdot\sin\tfrac\gamma2=\tfrac12|MA|$ hence $\phi=\frac{1}{2\sin\tfrac\gamma2}$ and

\begin{align} |CD|&=\frac{|AB|}{2\sin\tfrac\gamma2}=\mathbf{const}, \end{align} given a fixed chord $AB$.

Answer 2

Let $|\text{arc}(AB)|$ be the length of the arc $AB$.

From $\angle{BAC}=\angle{CAP}$, $$|\text{arc}(BC)|=|\text{arc}(CP)|\tag1$$

From $\angle{CAP}=\angle{PAD}$, $$|\text{arc}(DP)|=|\text{arc}(AD)|\tag2$$

From $(1)(2)$ we have $$\begin{align}|\text{arc}(CD)|&=|\text{arc}(CP)|+|\text{arc}(DP)|\\&=\frac 12|\text{arc}(BP)|+\frac 12|\text{arc}(AP)|\\&=\frac{1}{2}\left(|\text{arc}(BP)|+|\text{arc}(AP)|\right)\\&=\frac{1}{2}|\text{major arc(AB)}|\end{align}$$

Hence, it follows from this that $CD$ has a fixed length.