Dear All My Friends,

Please try one interesting problem: first prove following small lemma then use the lemma to prove Pythagorean theorem!

Lemma:
A, B, C are three points on one line by this order. D, E are two points on one side with respect to line contains A, B, C and AD//BE and BD//CE. F is midpoint of AC.
Result:
Area(DEF) = (Area(ABD) + Area(BCE))/2

Best regards,
Bui Quang Tuan

Lemma is pretty simple:

(FDE) = ((ADE) + (CDE))/2.

(ADE) = (ABD) and (CDE) = (BDE).

Yes, Alex!
And there is also one another very simple fact (in CTK) which combine with the lemma to generate Pythagorean theorem! :))
Could you find it?

Dear All My Friends,

Here is my solution for the problem.

Proof of lemma
AD//BE => Area(ABD) = Area(ADE)
BD//CE => Area(BCE) = Area(CDE)
Now we calculate areas of three triangles ADE, CDE, FDE. Denote ha, hc, hf as their altitude segments to common base DE respectively. These three altitude segments are parallel and all together in one side with respect to line A, B, C because D, E all together in one side with respect to line A, B, C. Since F is midpoint of AC and B is on segment AC, we have: hf = (ha + hc)/2. It means:
Area(DEF) = (Area(ADE) + Area(CDE))/2
or
Area(DEF) = (Area(ABD) + Area(BCE))/2
The proof is completed!
Note: if hb is altitude length of triangle BDE from B, using similarities:
ha/hb = AD/BE = BD/CE = hb/hc or hb^2 = hh*hc. It means:
Area(ADE)*Area(CDE) = Area(BDE)^2
We have two types of means of areas here but only can use arithmetic mean to prove Pythagorean theorem.

Proof of Pythagorean theorem
Suppose ABC is right triangle at C with side lengths a, b, c. It means AB is diameter of one circle, say (O) centered at O with radius c/2. Denote M as midpoint of arc ACB.
Construct a circle (Oa) centered at A passing C
Construct a circle (Ob) centered at B passing C
Line CM intersects (Oa) again at A'
Line CM intersects (Ob) again at B'

On the circle (O): arc(AM) = arc(BM) = 90 degree hence angle(ACA') = angle(BCB') = 45 degree. It means two triangles CAA', CBB' are right isosceles triangles at A and B respectively.
Now we use result in:
http://www.cut-the-knot.org/Curriculum/Geometry/TwoIntersectingCircles.shtml
with two circles (Oa), (Ob), line A'CB' and circle (O). From this, we have M is midpoint of A'B'.
Now we can use our lemma and calculate areas of triangles:

Area(AMB) = (Area(CAA') + Area(CBB'))/2
MO*AB/2 = (AA'*AC/2 + BB'*BC/2)/2
c/2*c/2 = (a*a/2 + b*b/2)/2
c^2 = a^2 + b^2
The proof is completed!

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.

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.

Yes, thank you. The small lemma of yours appears a powerful tool. It's all very interesting, but I am leaving tomorrow and am afraid I shall not be able to attend to this until our return on the 31st.

Actually what you did is proving Bottema's theorem along the way.

With best wishes,
Alex

Of course, I agree with you!
Best wishes and happy travel for you!
Bui Quang Tuan
>Yes, thank you. The small lemma of yours appears a powerful
>tool. It's all very interesting, but I am leaving tomorrow
>and am afraid I shall not be able to attend to this until
>our return on the 31st.
>
>Actually what you did is proving Bottema's theorem along the
>way.
>
>With best wishes,
>Alex