Geometry problem

• Dec 20th 2005, 05:56 AM
thduong711
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.
• Dec 20th 2005, 09:35 AM
CaptainBlack
Quote:

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 $\displaystyle A$ is the
orthocentre of triangle $\displaystyle PQR$ it is sufficient to prove that $\displaystyle QA$ extended
is a normal to $\displaystyle PR$, and that $\displaystyle RA$ extended is a normal to $\displaystyle PQ$.

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

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

==================================================

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

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

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

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

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

and

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

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

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

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

RonL
• Dec 20th 2005, 09:21 PM
thduong711
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