Perpendiculars from the orthocentre to angle bisectors

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Perpendiculars from the orthocentre to angle bisectors

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Source: South African MO 2008 Q5

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djb86

#1 h May 27, 2012, 7:53 AM • 2 Y

Y by Adventure10, Mango247

Triangle $ABC$ has orthocentre $H$ . The feet of the perpendiculars from $H$ to the internal and external bisectors of $\hat{A}$ are $P$ and $Q$ respectively. Prove that $P$ is on the line that passes through $Q$ and the midpoint of $BC$ . (Note: The ortohcentre of a triangle is the point where the three altitudes intersect.)

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#2 h May 27, 2012, 5:54 PM • 1 Y

Y by Adventure10

Nice, simply stated problem. We should make a collection of these.

Let $G$ , $E$ , and $F$ be the feet of the perpendiculars from $A$ , $B$ , and $C$ to the opposite sides of the triangle, and let $M$ be the midpoint of $BC$ and $R$ be the midpoint of $AH$ . Since $\angle AQH= \angle AFH =\angle APH= \angle AEH= 90$ , $AQFHPE$ is cyclic with circumcenter $R$ (which lies on $QP$ since $QAPH$ is a rectangle), and $RF=RE$ . Since $\angle FAP= \angle EAP$ , $PF=PE$ so $RFPE$ is a kite. Also, we know $FREMG$ is cyclic as the points lie on the nine-point circle. Since $\angle RGM=90$ , $RM$ is a diameter, so $R$ , $P$ , and $M$ are collinear, so $Q$ , $P$ , and $M$ are collinear.

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r1234

#3 h Jun 3, 2012, 2:23 PM • 2 Y

Y by Adventure10, Mango247

Let $A'$ be the midpoint of $AH$ .Easy to check that $\angle QA'A=|B-C|$ .Again if $M$ is the midpoint of $BC$ , $A'M$ is the diameter of the nine-point circle of $\triangle ABC$ .So $\angle HA'M=|B-C|$ .Hence $PQ$ passes through $M$ .

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