As shown in the figure, in the right triangle ABC, ∠ C = 90 ° and BC is bisected by points D and E, and BC = 3aC, then ∠ AEC + ∠ ADC + ∠ ABC=______ °．

As shown in the figure, in the right triangle ABC, ∠ C = 90 ° and BC is bisected by points D and E, and BC = 3aC, then ∠ AEC + ∠ ADC + ∠ ABC=______ °．

As shown in the figure, bcmn is a square with BC as the edge, and points P and Q are taken on Mn to make MP = PQ = QN. If AP, PD and DQ are connected, then am = CD, MP = AC, △ ACB ≌ △ PQD can be obtained easily, so AP = ad and ∠ DAC + ∠ PAM = 90 °, i.e. △ APD is isosceles right triangle, so ∠ ABC + ∠ ADC = ∠ PDQ + ∠ ADC =

As shown in the figure, in △ ABC, ∠ C = 90 °, D, e are two points on the edge of BC, and ∠ ABC = 12 ∠ ADC = 13 ∠ AEC, BD = 11, de = 5 are known, and AC length is calculated

If ∵ - ABC = 12 ∵ ADC = 13 ∵ - AEC, ∵ - ABC = ∵ DAE, and ∵ - AEB = ∵ DEA (common angle), ∵ ade ∽ BAE, then Abba = aebe = DEAE. From ∵ abd = ∵ bad, we get ad = BD = 11, be = BD + de = 16. ∵ AE2 = be · de = 16 × 5 = 80, AE = 45, ab = ad · Aede = 4455