I can see a concurrency(Self-made)
- Fm Jakaria
- Posts:79
- Joined:Thu Feb 28, 2013 11:49 pm
In a triangle ABC, let the A-excircle,B-excircle,C-excircle respectively touch the sides BC,CA,AB at X,Y,Z. Prove that the respective perpendiculars from A to YZ,B to ZX, C to XY are concurrent.
You cannot say if I fail to recite-
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.
- Fatin Farhan
- Posts:75
- Joined:Sun Mar 17, 2013 5:19 pm
- Location:Kushtia,Bangladesh.
- Contact:
Re: I can see a concurrency(Self-made)
$$AB,BC,CA$$ are tangent to the three circles
$$\Rightarrow BX^2+CY^2+AZ^2=XC^2+YA^2+ZB^2$$
Now using the facts,
$$(a) X,Y,Z \text{ are points on the sides } BC,CA,AB \text{ of } \triangle ABC.\text{ Then the
perpendiculars to}$$ $$\text{ the sides at these points meet in a common point } P \text{ if and only if }$$ $$
BX^2+CY^2+AZ^2=XC^2+YA^2+ZB^2$$
$$(b)$$ In $$\triangle ABC$$ if $$X ,Y,Z \text{ are the feet of the perpendiculars from the point } P$$ $$\text{ to the sides }BC,CA,AB \text{ respectively. Then the perpendiculars from } A,B,C$$ to $$YZ,ZX,XY
\text{ respectively are concurrent.}$$
$$\Rightarrow BX^2+CY^2+AZ^2=XC^2+YA^2+ZB^2$$
Now using the facts,
$$(a) X,Y,Z \text{ are points on the sides } BC,CA,AB \text{ of } \triangle ABC.\text{ Then the
perpendiculars to}$$ $$\text{ the sides at these points meet in a common point } P \text{ if and only if }$$ $$
BX^2+CY^2+AZ^2=XC^2+YA^2+ZB^2$$
$$(b)$$ In $$\triangle ABC$$ if $$X ,Y,Z \text{ are the feet of the perpendiculars from the point } P$$ $$\text{ to the sides }BC,CA,AB \text{ respectively. Then the perpendiculars from } A,B,C$$ to $$YZ,ZX,XY
\text{ respectively are concurrent.}$$