In square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 °, try to explain s △ AEF = s △ Abe + s △ ADF

Extending FD to h, making HD = be, connecting ah, it is easy to prove that ah = AE, the area of triangle ahd is equal to the area of triangle Abe, because the angle EAF = 45 degrees, it is easy to prove that the angle HAF = 45 degrees, so triangle AEF and triangle AHF are congruent, that is, the area is also equal,

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1. It is known that the quadrilateral ABCD is a square with a side length of 6 cm. The area of the triangle ECF is 3 square cm larger than that of the triangle ADF
2. As shown in the figure, it is known that the quadrilateral ABCD is a square with a side length of 5cm. The area of the triangle ECF is 5 square centimeters larger than that of the triangle ADF. Find the length of the line CE
3. If the areas of △ CEF, △ Abe and △ ADF are 3, 4 and 5 respectively, the area of △ AEF can be obtained
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12. As shown in the figure, in square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 ° to prove s △ AEF = s △ Abe + s △ ADF
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