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.

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    $\begingroup$ To aid people who might be interested in helping, can you please provide a drawing? $\endgroup$ – Kim Jong Un Oct 20 '14 at 13:58

This is complicated and I have to break it into 3 parts.

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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: $\theta = 180^0 – (\phi + \beta) - \alpha = 180^0 – 90^0 - \alpha = 90^0 - \alpha = 90^0 - \beta = \phi$

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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 $OHZH’$.

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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.]

  • $\begingroup$ This is obvious - To whom ? Also, regarding your entire auxiliary construction from facts $2$ and $3$, it is unclear how exactly all the extra points are constructed. $\endgroup$ – Lucian Dec 31 '15 at 15:49
  • $\begingroup$ @Lucian A proof has been added. I am happy to provide further explanation if any one of the construction is not valid. $\endgroup$ – Mick Dec 31 '15 at 18:33

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