If it is a quadrilateral, it cannot be proved. If it is a square, then BC=CD=AD, ∵F is the midpoint of CD, then FD=CF= 1/2CD, then FD =1/2bc, ∴ CE =1/2fd is CE: FD. ∴CF:AD= 1/2, ∴CF:AD=CE:FD and ∠ d = ∠ c = 90, then △CEF is similar to △DFA, ∴ CFE = ∠ DAF.

Solution: extend AD to E to make AD=DE, and connect be≈ADC =∠b dead = debd = DC ∴△ ADC △ EDB ∴ Be = AC = 3 ∫ AB = 5ae = 2ad = 2×.

According to Pythagorean theorem AB squared AC squared =BC squared, and according to triangle area AD multiplied by BC=AC multiplied by AB, the equations AC=3 and BC=5 are solved.