Geometry problem

Fast Help!

I have to complete my homework in 2 days and this problem I found most difficult:

Let ABC be a right triangle with hypotenuse BC. Suppose that M is the midpoint of BC and H is the feet of the perpendicular dropped from A onto BC. A point P, distinct from A, is chosen on the opposite ray of ray AM. Let the line through H perpendicular to AB intersect PB at Q; and let the line through H perpendicular to AC meet PC at R. Prove that A is the orthocenter of triangle PQR.

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Originally Posted by thduong711

I have to complete my homework in 2 days and this problem I found most difficult:

Let ABC be a right triangle with hypotenuse BC. Suppose that M is the midpoint of BC and H is the feet of the perpendicular dropped from A onto BC. A point P, distinct from A, is chosen on the opposite ray of ray AM. Let the line through H perpendicular to AB intersect PB at Q; and let the line through H perpendicular to AC meet PC at R. Prove that A is the orthocenter of triangle PQR.

See diagram. By definition of the Orthocentre to prove that $A$ is the
orthocentre of triangle $PQR$ it is sufficient to prove that $QA$ extended
is a normal to $PR$ , and that $RA$ extended is a normal to $PQ$ .

Now it is sufficient to prove that in general $QA$ extended is a normal $PR$ ,
as by an equivalent argument $RA$ will be a normal to $PQ$ .

Now the best way to proceed as far as I can see is to introduce
coordinates, and use coordinate geometry to show that angle $QSR$
in the diagram is a right angle.

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<<You will need to check the algebra in this carefully>>

Let $A$ be the origin $(0,0)$ , $B$ be $(0,b)$ and $C$ be $(c,0)$ .

Then $M$ is $(c/2,b/2)$ .

The slope of $CB$ is $-b/c$ , so the slope of $AH$ is $c/b$ , in fact $AH$
is the line $y=(c/b)x$ . The line $CB$ has equation $y=(-b/c).x+b$ ,
so $H$ is the point of intersection of:

$y=(c/b).x$ ,

and

$y=(-b/c).x+b$ .

Which is the point $(b^2.c/(b^2+c^2), b.c^2/(b^2+c^2))$ .

Now $P$ lies on $AM$ , and may be written as $-\mu.(c/2,b/2)$ for some
$\mu>0$ .

To finish find the equation on the line through $P$ and $B$ , and from that
find the coordinates of $Q$ . Find the slope of the line through $Q$ and $A$ ,
which should be minus the slope of the line through $P$ and $C$ .

RonL

Sorry,but do you have another solution that just using similar/congruent triangles, circles...etc...I haven't learn about your way to solve this, so I don't understand that much, and I think my teacher has another more simple solution