Let $ABC$ be a non-isosceles triangle and $I$ be the intersection of the three internal angle bisectors. Let $D$ be a point of BC such that $ID\perp BC$ and $O$ be a point on AD such that $IO\perp A$D . Prove $OD$ is the angle bisector of the angle BOC.
This is complicated and I have to break it into 3 parts.
Fact#1) In the figure, $BOU$ is a straight line and is composed of angles at $O$ with $\alpha + \beta + \phi + \theta = 180^0$. It is given that $\angle IOA = \beta + \phi = 90^0$. If $\alpha = \beta$, then $\phi = \theta$. This is obvious and therefore the proof is skipped.
EDIT: Here is a simple proof: $\theta = 180^0 – (\phi + \beta) - \alpha = 180^0 – 90^0 - \alpha = 90^0 - \alpha = 90^0 - \beta = \phi$
Fact#2) Circles $OHZ$ and $OVU$ touch each other internally at $O$. $XO$ and $XZ$ are tangent pairs to the circle $OHZ$. $XZ$ cuts the circle $OUV$ at $U$ and $V$. Then, $OZ$ bisects $\angle VOU$ (i.e. $\alpha = \beta$.
$\lambda + \alpha = \mu$ [tangent properties]
$= \beta + \angle V$ [ext. angle of triangle]
$= \beta + \lambda$ [angles in alternate segment]
∴ $\alpha = \beta$
Initial construction: 1) Through $O$, draw $XOX’$ parallel to $AB$; where $X’$ is on $BC$.
2) Through $O$, draw $YOY’$ perpendicular to $XOX’$; where $Y$ is on $AB$.
3) Extend $OI$ to cut $AC$ at $Z$.
4) Construct the perpendicular bisector of $OZ$ so that it cuts $X’OX$ at $X$ and $YOY’$ at $K$.
5) Using $K$ as center & $KO$ as radius, draw the circle $OHZH’$ (in red); $H$ & $H’$ are arbitrary points (but on the opposite sides of $OZ$ of the circle.
Through the above, we have successfully created $XO$ and $XZ$ as a pair of tangents to the circle $OHZH’$.
Additional construction: 6) Extend $BO$ to cut $XZ$ at $U$ and extend $XUZ$ to cut $OC$ at $V$.
7) Let the perpendicular bisector of $UV$ cut $YY’$ at $J$.
8) Using $J$ as center & $KO$ as radius, draw the circum-circle $OUV$ (in blue).
Based on the above construction, $OZ$ bisects $\angle UOV$ (i.e. $\alpha = \beta$.) [See fact #2.]
Therefore, $\theta = \phi$. [See fact #1.]
