(1) extends the intersection of AD and BC at point H, so that the parallel lines of BC pass through point D, and the result is 1 16.

Solution: Make DF∨BC intersect AB at point F, ∠B =∠ and DFA CD = BF = 10, DF = BC. ∴AF = 50- 10 = 40 .

∠A +∠B = 90 ∴∠A +∠DFA = 90 .

∫ COSA = AD/AF = 4/5 ∴ AD = 4/5× AF = 4/5× 40 = 32. ∴ DF = 24 = BC ∴ Peripheral = 32+10+24+50 =

(2) whether the crossing point n is NE⊥AB of point e, Mn cos nma = me = y；; Because COSA = AE/AN = 4/5, AE = 4/5X. AM =？ AB = 25. So y = 25-4/5x.

(3) The extension line AC CB of figure NM is at point P. 。

Solution: Extend the intersection of AD. BC is at point h, connecting HM. ≈ A+≈ B = 90 ∠ H = 90. ∫ The midpoint of AB, ∴MH =? AB = BM = AM = 25, ∴∠ A = ∠ AHM = ∠ P ∠ B = ∠ BHM. ∴∠ P+∠ BHM = 90 ∴△ PHMRT. Got (1): AD = 32, BC = 24.

∫DH/ah = ch/BH = CD/ab = 10/50 = 6∴dh = ch∴ah = 40∫△HCD∽△MHP∽△HNP

∴ ch/CD = hm/ph ∴ 6/10 = 25/ph ∴ ph = 125. ∫ch/DH = NH/ph∴6/8 = nh/ 125/4∴nh = 125/4∴an = ah-NH = 40- 125/4 = 35/4 .