That is, when E and F are in AC and CB, the area of quadrangle ECFD is half that of triangle ABC, and it is more than half that of triangle after exceeding it.

Proof: In Figure 2, < mde = < NDF is formed because they are rotating.

Since d is the midpoint of the hypotenuse, DM=DN.

Because the two angles and the same sides of the two angles are equal, the triangle DME is congruent with the triangle DNF.

That is, S△DME=△DNF.

Then the area of quadrangle ECFD is equal to the area of square MCND, and the area of square MCND is half that of triangle ABC, which is proof.

S△DEF+S△CEF= 1/2S△ABC

In figure 3, because: DC, similarly triangle DBF, DF> dB.

That is DE * DF & gtCD * DB & gtAB*CD/2.

Namely: s △ def > 1/2S△ABC

Then S△DEF+S△CEF= 1/2S△ABC does not hold.