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Simson/Wallace Line Proof

Date: 11/19/2002 at 12:27:00
From: Mukesh Kumar
Subject: Geometry

From a point P on the circumcircle of the triangle ABC perpendiculars
are dropped to the sides AB, BC, CA. Prove that the line joining the 
feet of the perpendiculars (the Simson line or the pedal of the point 
p with respect to triangle ABC) bisects the line joining the 
orthocentre of triangle ABC and point P.

Please send me a solution. I have tried my level best, but I can't 
solve it.

Thank you.

Date: 11/20/2002 at 08:44:48
From: Doctor Floor
Subject: Re: Geometry

Hi, Mukesh,

Thanks for your question.

Let's give it a try. If you want to follow this proof, you'll need 
pencil and paper to draw the needed sketches with me. 

We start with triangle ABC and its circumcircle and we construct 
orthocenter H. Without loss of generality the point P at the 
circumcircle is between A and C (if P is A, B, or C, a simple case 
remains). We draw the Simson/Wallace line l that passes BC in D, and 
AC in E (the point at AB I won't use).

Draw the line AH, giving the intersection points A' at BC and H' at 
the circumcircle. Now draw H'P, giving intersection points F at BC 
and G at l. Finally draw HP with intersection point S at l. That was 
a whole lot of drawing. 

For your reference, here is a figure:


Now we are going to show that FG = GP (*) and that FH//l (//: is 
parallel to) (**). If we have shown (*) and (**) we can easily 
conclude that HS = SP, which is what we want to prove.

To show (*) and (**) we are going to show some angle equalities. First 
consider triangles CDP and CEP. They're both rectangular and have CP 
as joint hypotenuse. This means there is a circle passing through all 
four vertices of the two triangles. But this also means that 
angle(EDP) = angle(ECP) since they intercept the same arc on this 

Also, angle(ECP) = angle(AH'P) since they intercept the same arc on 
the circumcircle.

Now we have to recall the property of the nine point circle, being the 
productfigure of the circumcircle with factor 1/2 and productcenter 
H. The nine point circle passes through A'; thus A'H = A'H'. Then it 
is easy to see that triangles H'A'F and HA'F are congruent, and 
angle(AH'P) = angle(A'H'F) = angle(A'HF).

Finally angle(AH'P) = angle(H'PD) because of being "Z-angles" (AH = 
AH' and PD are parallel since they are both perpendicular to BC).

Concluding, we have:

  angle(EDP) = angle(ECP) = angle(AH'P) = angle(A'HF) = angle(H'PD)

From angle(EDP) = angle(H'PD) we can conclude that PG = DG, but in the 
rectangular triangle FDP then also angle(GDF) = angle(GFD) hence 
FG = DG. This means that FG = GP (*).

From angle(EDP) = angle(A'HF) we can conclude that FH//l (**).

Having shown that (*) and (**) hold, we have proven that the 
Simson/Wallace line passes halfway between P and H.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum

Date: 11/21/2002 at 08:52:11
From: Mukesh Kumar
Subject: Thank you (Geometry)

Hi, Dr. Math,

Thank you for the excellent proof you have given me. 

Mukesh Kumar
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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