Angle Bisector in Circle

Source

Angle Bisector in Circle, source

Problem

Points $A,B,C,D\,$ on a circle satisfy $\angle BAD=\angle CAD.$

Angle Bisector in Circle, illustration

If $M\,$ is the midpoint of $AD,\,$ prove that

$BM+CM\ge AD.$

Solution

We assume that the circumcenter $O\,$ of $\Delta ABC\,$ is the center of the coordinate system, that the circumradius equals $1,\,$ and that the triangle is oriented counterclockwise, i.e., positively. Let points $A,B,C\,$ relate to complex numbers $a^2,b^2,c^2.\,$ Then $D=bc\,$ and $M=\displaystyle \frac{a^2+bc}{2}.\,$ We have to prove that

(1)

$\displaystyle\begin{align} MB+CM&=\left|b^2-\frac{a^2+bc}{2}\right|+\left|c^2-\frac{a^2+bc}{2}\right|\\ &\ge |a^2-bc|=AD. \end{align}$

Now, $\displaystyle \frac{b^2}{a^2}=\cos 2C+i\sin 2C,\,$ $\displaystyle \frac{a^2}{c^2}=\cos 2B+i\sin 2B,\,$ $\displaystyle \frac{c^2}{a^2}=\cos 2B-i\sin 2B,\,$ $\displaystyle \frac{a}{c}=\cos B-i\sin B.\,$ The inequality (1) is then equivalent to

$\displaystyle \left|\frac{b^2}{a^2}-\frac{1}{2}\left(1+\frac{b}{a}\cdot\frac{c}{a}\right)\right|+\left|\frac{b^2}{a^2}-\frac{1}{2}\left(1+\frac{b}{a}\cdot\frac{c}{a}\right)\right|\ge\left|1-\frac{b}{a}\cdot\frac{c}{a}\right|,$

or,

$\displaystyle\begin{align} &\left|\cos 2C+i\sin 2C-\frac{1}{2}\left(\cos (C-B)+i\sin (C-B)\right)\right|\\ &\qquad\qquad+\left|\cos 2B-i\sin 2C-\frac{1}{2}\left(\cos (C-B)+i\sin (C-B)\right)\right|\\ &\qquad\ge |1-(\cos (C-B)+i\sin (C-B))|. \end{align}$

Further

$\displaystyle\begin{align} &\left|\cos 2C+i\sin 2C-\frac{1}{2}\left(\cos (C-B)+i\sin (C-B)\right)\right|\\ &\qquad\qquad=\sqrt{\left(\cos 2C-\frac{1}{2}\cos (C-B)\right)^2+\left(\sin 2C-\frac{1}{2}\sin (C-B)\right)^2}\\ &\qquad\qquad=\sqrt{1+\frac{1}{4}-\cos [2C-(C-B)]}=\sqrt{\frac{5}{4}+\cos A}. \end{align}$

Similarly,

$\displaystyle \left|\cos 2B-i\sin 2C-\frac{1}{2}\left(\cos (C-B)+i\sin (C-B)\right)\right|=\sqrt{\frac{5}{4}-\cos A}$

and

$\displaystyle\begin{align} |1-(\cos (C-B)+i\sin (C-B))| &= \sqrt{\left(1-\cos (C-B)\right)^2+\left(\sin (C-B)\right)^2}\\ &=\sqrt{2-2\cos (C-B)}=2\sin\left(\left|\frac{C-B}{2}\right|\right). \end{align}$

Now,

$\displaystyle\begin{align} \left(\sqrt{\frac{5}{4}+\cos A}+\sqrt{\frac{5}{4}-\cos A}\right)^2 &= 2\cdot\frac{5}{4}+2\sqrt{\left(\frac{5}{4}\right)^2-\cos^2A}\\ &\ge\frac{5}{2}+2\sqrt{\left(\frac{5}{4}\right)^2-1}=\frac{5}{2}+2\frac{3}{4}=4. \end{align}$

Thus, $\displaystyle \left(\sqrt{\frac{5}{4}+\cos A}+\sqrt{\frac{5}{4}-\cos A}\right)^2\ge 4.\,$ Consequently,

$\displaystyle \sqrt{\frac{5}{4}+\cos A}+\sqrt{\frac{5}{4}-\cos A}\ge 2\ge 2\sin\left(\left|\frac{C-B}{2}\right|\right),$

as required.

Acknowledgment

The problem has been posted by Miguel Ochoa Sanchez. Marian Dinca has communicated to me the problem, along with his solution.

 

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