Here is an outline of a proof we skipped in class. We prove that,
for any two triangles Δ ABC and Δ A′B′C′,
we have
A(Δ A′B′C′)
A(Δ ABC)

=

D(Δ A′B′C′)
D(Δ ABC)

.


(*)

In the proof, we'll be using the following fact, often called the
comparison theorem:
Suppose x > m/n if and only if y > m/n for any fraction
m/n. Then x=y.
Before we start, you should convince yourself that this is true.
(The idea is simply that this means x and y are bounded 
above and below  by the same rational numbers. If they were
different, there would be a rational number between them. Of
course, proving this statement is a little bit offtopic,
but you should think about it.)
We begin the proof by noting that if D(Δ ABC) = D(Δ
A′B′C′), then A(Δ ABC) = A(Δ
A′B′C′) by Bolyai's theorem. So we'll assume
that D(Δ ABC) > D(Δ A′B′C′) (if
they're different, then one of them must be larger).
We choose Q on BC so that D(Δ ABQ) = D(Δ
A′B′C′) (to see the picture, please see the pdf file). Space the
points P_{1}, P_{2}, and so on so that there are n
triangles Δ AP_{k}P_{k+1}, each with the
same defect, namely D(Δ AP_{k} P_{k+1}) =
(1/n) D(Δ ABC). (These points are not presumed to be evenly
spaced along BC.) Pick m; we'll show that

m
n

<

A(Δ A′B′C′)
A(Δ ABC)

if and only if

m
n

<

D(Δ A′B′C′)
D(Δ ABC)



which, by the comparison theorem, is enough to prove
equation (*).

Show that
m
n

<

A(Δ A′B′C′)
A(Δ ABC)

if and only if

m
n

A (Δ ABC) < A( Δ ABQ).



Show that for any k (and in particular for k=m)
A(Δ ABP_{k}) < A(Δ ABQ)
if and only if
D(Δ ABP_{k}) < D(Δ ABQ).



Explain why (m/n) D(Δ ABC) = D(Δ ABP_{m})
and (m/n) A(Δ ABC) = A(Δ ABP_{m}).
Hint: Use Bolyai's theorem which, in this case, you can
take to say that if the defects are equal, the areas are equal.

Combine the above to show equation (*).