I can see a concurrency(Self-made)

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Fm Jakaria
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I can see a concurrency(Self-made)

Unread post by Fm Jakaria » Sun Aug 24, 2014 5:45 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,
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whether I may, drown in tub and die.

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Fatin Farhan
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Re: I can see a concurrency(Self-made)

Unread post by Fatin Farhan » Tue Aug 26, 2014 11:18 am

$$AB,BC,CA$$ 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 $$YZ,ZX,XY
\text{ respectively are concurrent.}$$

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