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20 13 the last question of mathematics in the senior high school entrance examination in Hongkou district.
(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 .