The seventh question, don't integer so many angles, connect AD directly, then prove two parallelograms, and then you should be able to prove the eighth question of AB=DE. Use parallel lines with the same internal angle and diagonal, and then divide them equally with the two diagonal lines of parallelogram to get congruent triangle, and you can know that it is the ninth question of OE=OF, ∠ 1+∠3=90. Add a square to find a pair of equilateral sides, and the angle can prove congruent triangles, which can prove the tenth problem of DE=DF, because ∠ 1=∠B, so DE = BE because ∠C=2∠B, ∠ AED = ∠1. And because AD is the bisector of △ABC, ∠ CAD = ∠ DAE; AD is the public party, so △ ACD △ DAE; So AC=AE, CD=DE=BE, so AB=AC+CD.