prove that altitude of a triangle r concurrent no

I can't make sketch here to show the proof clearly but
I'll describe the sketch so that you can follow by making the sketch on your paper while
reading this.

Let's draw a triangle with sides
a, b,
c. Pls. put c on the longest of the three sides. Now name
the vertices as A, B,
C with A opposite side
a; B opposite side
b and C opposite
side c.

Now draw the first
altitude perpendicular to side c.. This will pass through
vertex C. We will call this as
"altC". Put a small square at the intersection of
altC and side c, to indicate
right angle and mark this intersection as point
F.

Next draw the second
altitude perpendicular to side a. This line will pass
through vertex A. We will call this as
"altA". Put a small square at the intersection of
altA and side a, to indicate
right angle and mark this intersection as point
D. (Note: If angle C
is acute, then point D is on side a in the triangle,
else point D is on the extension of side
a outside the
triangle.)

Finally draw the last altitude
through vertex B and perpendicular to side
b and we will call this as
"altB". Mark the intersection of
altB and side b as point
E and put small square on it. (Again if angle
C is acute, then point E is on
side b in the triangle, else point E is on the extension of
side b outside the
triangle.)

Now locate the intersection of
altA and altC and mark this as
point O1, then locate also the intersection of
altB and altC and mark as
point O2. then locate also the intersection
of altA and altB and mark as
point O. (For your guide if
angle C is acute, then point O1,
O2 and O is inside the triangle,
else point O1, O2
and O are
outside the triangle.)

If
your drawing is perfect the you'll see that the altitudes are concurrent. But it's
better if they're not. so that we a triangle of error
O1O2O

O1,
O2 and O are the same point, if they are
concurrent and FO1=FO2. So we can make a proposition
that:

The altitudes of a triangle
are concurrent because distance FO1 and distance FO2 are
equal.

Proof:

From
right triangle AFC

FC = bsinA;     AF =
bcosA

From right triangle BFC

FC = asinB;     BF = acosB

bsinA = asinB

b = asinB/sinA

But
<ABD = 90-A, thus <FO2B = A

Also <BAE
= 90-B, thus <FO1A = B

From right triangle 
AFO1

tanFO1A = AF/FO1

FO1 =
AF/ tanFO1A

FO1 = bcosA/tanB

FO1 = (asinB/sinA)cosA /tanB

FO1
=acotAcosB

From right triangle 
BFO2

tan FO2B = BF/FO2

FO2 =
BF/tan FO2B

FO2 = acosB/tanA

FO2 = acosBcotA

Distance FO1 is equal to distance FO2,
therefore

The altitudes of a
triangle are concurrent