# A One-Sided Inequality in Triangle

### Proof

In an acute triangle, the angle bisector, the altitude, and the symmedian from the same vertex, all fall on the same side from the midpoint of the opposite side.

For this reason, we have

$\displaystyle\begin{array}{ccc} BA'\lt\frac{a}{2}, & \;CB'\lt\frac{b}{2}, & \;AC'\lt\frac{c}{2},\\ BA''\lt\frac{a}{2}, & \;CB''\lt\frac{b}{2}, & \;AC''\lt\frac{c}{2},\\ BA'''\lt\frac{a}{2}, & \;CB'''\lt\frac{b}{2}, & \;AC'''\lt\frac{c}{2}. \end{array}$

Multiplying the rows and adding the results yields the required inequality.

To clarify the inequalities, we make a couple of observations.

First, if $M\;$ is the midpoint of $AC,\;$ then $A'\;$ lies between $A''\;$ and $M.$

Second, if $D\;$ is the midpoint of the arc $BC,\;$ opposite $A\;$ and $E\;$ is the second intersection of $AM\;$ with the circumcircle $(ABC),\;$ then $\angle A'AA'''=\angle DAE\lt\angle CAD=\angle BAD,\;$ implying that $A'''\;$ is between $B\;$ and $M.$

### Acknowledgment

The problem has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. The problem came from his book Math Storm.