Rotate △CDF by 180 degrees with d as the rotation center, so that CD and BD coincide. F falls in f' because ∠ EDF = ∠ EDF' = 90 degrees ED = EDDF = DF', so △ DEF △ DEF' because ∠B=∠C=45 degrees, so ∠ ABF' = 90 degrees in Rt△EBF'' Let AB = 2 = 13 2 is solved to get the solution of X = 655: if point c intersects CF//BD and the extension line of AB is at F, then AC⊥CE and quadrilateral BFCD are parallelograms ∴ CF = BD = AC = 6 ∴ AF =10 ∴ RT. If m and b coincide, then AH=BG=BN△BHG and △BGN are equilateral triangles. What is quadrilateral MNGH? What is a quadrilateral diamond? AD=2= 1/2BC△ABC is a right triangle. Let two right-angled sides be A and B, then AB = 3 √ 7 = > (ab) 2 = a22ab =166 √ 7a2b2 = bc2 =16area s =1/2ab = (3/2) √ 71,Solution: Because the isosceles right triangles ABC and AD are the median lines, Because the isosceles right triangle ABC and AD are the midline and AD is the bisector of BAC, EAD= 1/2BAC=FCD=45 degrees. So the triangle ADE is equal to the triangle CDF. So EA=FC=5, because AB=AC, AF= 12, so EF= 13, and triangle DEF is an isosceles right triangle, and its area = base EF* height/2 =13 * (13/2).