Dear Alex,

Of course, you are right! There are some ways to get midpoint in order to use area lemma and we can also directly prove because the proof very short.
I only want to combine some facts already in CTK so readers can remember as much as possible.
It is also my way when I need one midpoint, I think about "two intersecting circles".
This proof is very close with "broken chord" proof in which we use mean of two lengths and here we use mean of two areas.

The area lemma is also can be use for following problem:
To find point X on segment side BC of triangle ABC such that area of XB'C' is maximum, here: XB'C' is triangle inscribed in ABC:
B' is on AC and XB'//AB
C' is on AB and XC'//AC
After proof we can see that for all three segments BC, CA, AB, only median triangle is a maximum area triangle.

Proof:
For short, we denote:
S = area(ABC)
S1 = area(XBC')
S2 = area(XCB')
S3 = area(XB'C') = area(AB'C')

By our lemma: S3 = Sqrt(S1*S2)
S = S1 + S2 + S3*2 = S1 + S2 + Sqrt(S1*S2)*2
=(Sqrt(S1) + Sqrt(S2))^2
4*S3 = (Sqrt(S1) + Sqrt(S2))^2 - (Sqrt(S1) - Sqrt(S2))^2
= S - (Sqrt(S1) - Sqrt(S2))^2
Hence S3 maximum iff S1 = S2 and we get XB'C' is median triangle

Best regards,
Bui Quang Tuan

>Very good.
>
>There, perhaps, is a shortcut via Bottema's theorem.
>
>On sides AC and BC of right (at C) ΔABC form squares
>and join the corner opposite to C, say, A' in one square and
>B' in the other. The midpoint M of A'B' does not depend on
>the position of C and is the vertex of the right isosceles
>triangle AMB. You are in a position to apply your lemma
>right away.