It is known that in diamond ABCD, e and F are points on AB and ad respectively, and AE = AF. verification: CE = CF

It is known that in diamond ABCD, e and F are points on AB and ad respectively, and AE = AF. verification: CE = CF


It is proved that ∵ ABCD is rhombic, ∵ ad = AB = BC = CD, ∵ d = ∵ B, AF = AE, ∵ FD = EB, ≌ DFC ≌ bec (SAS), ≌ CE = CF



In parallelogram ABCD, extend BC to P, extend DC to Q, so that CP / BC = CQ / DC = m, when s △ Apq = s ABCD, then M=


If CP / BC = CQ / DC, angle BCD = angle PCQ, then triangle BCD is similar to triangle PCQ, then PQ / BD = m (1), suppose triangle BCD, height on BD side is h, triangle PCQ, height on PQ side is h, then H / h = m, that is, H = MH, because parallelogram ABCD, then triangle Apq, height on PQ side is 2H + mhsabcd = BD * H (2), s △ Apq = 1 / 2pq * (2H + MH