As shown in the figure, in RT △ AOB, the circle AB with radius OA intersects at point C. If Ao = 5, OB = 12, the length of BC is obtained

As shown in the figure, in RT △ AOB, the circle AB with radius OA intersects at point C. If Ao = 5, OB = 12, the length of BC is obtained

Through point E, make OE ⊥ AC at point E, ∵ AOB = 90 °, Ao = 5, OB = 12, ∧ AB = 13, ∧ EO × AB = Ao × Bo, ∧ EO = Ao × boab = 5 × 1213 = 6013, in RT △ AEO, AE = AO2 − EO2 = 2513, ∧ AC = 2513 × 2 = 5013, ∧ BC = 13-5013 = 11913

As shown in the figure, in RT △ AOB, the circle AB with radius OA intersects at point C. If Ao = 5, OB = 12, the length of BC is obtained

Through point E, make OE ⊥ AC at point E, ∵ AOB = 90 °, Ao = 5, OB = 12, ∧ AB = 13, ∧ EO × AB = Ao × Bo, ∧ EO = Ao × boab = 5 × 1213 = 6013, in RT △ AEO, AE = AO2 − EO2 = 2513, ∧ AC = 2513 × 2 = 5013, ∧ BC = 13-5013 = 11913

Known: as shown in the figure, in RT △ ABC, point O is on AB, the circle with o as the center and OA length as radius intersects with AC and ab at points D and e respectively, and ∠ CBD = ∠ A. (1) judge the position relationship between line BD and ⊙ o, and prove your conclusion; (2) if BC = 2, BD = 52, find the value of ADAO

(1) The straight line BD is tangent to ⊙ O. it is proved that as shown in Fig. 1, connecting OD. ∵ OA = OD, ∵ a = ⊙ ADO. ∵ C = 90 °, and ∵ CBD + ⊙ CDB = 90 °, and ∵ CBD = ⊙ a, ∵ ADO + ⊙ CDB = 90 °, and ∵ ODBC = 90 °, straight line BD is tangent to ⊙ O. (2) solution 1: as shown in Fig. 1, connecting de. ∵ C = 9

Given RT △ ABC and RT △ EBC, ∠ B = 90 °, with point o on edge AC as the center and OA as the radius, ⊙ O and EC are tangent to point D, ad ∥ BC. ∠ e = ∠ ACB

This problem is not difficult to solve,
∵ AD∥BC ∠ACB=∠OAD
OA=OD ∠ADO=∠OAD
∴∠ADO=∠ACB
In RT △ ade, e + ade = 90 degree
Od is tangent, OD ⊥ CE is ∠ ADO + ∠ ade = 90 degree
∴∠E=∠ADO=∠ACB