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As shown in the figure, it is known that △ ABC and △ DEB are isosceles right triangles, ∠ ACB = ∠ EDB = 90 °, point E is on side AC, CB and ED intersect at point F
It is proved that: (1) both △ ABC and △ DEB are isosceles right triangles, ∠ ACB = ∠ EDB = 90 °, ∠ Abe = ∠ CBD, ebbd = abbc = 22, ‖ △ Abe ∽ CBD; (2) the four points B, D, C and E are in common circle, ∠ CDE = ∠ CBE, ∠ CBD = ∠ Abe; ∵ △ ABC and △ DEB are equal
It is known that AOB and BOC complement each other, OD is bisector of AOB, OE is in BOC, BOE = 12 EOC, DOE = 72 ° and the degree of EOC is calculated
Let ∠ EOB = x, then ∠ EOC = 2x, then ∠ BOD = 12 (180 ° - 3x), then ∠ BOE + ∠ BOD = ∠ doe, that is, x + 12 (180 ° - 3x) = 72 ° and the solution is x = 36 ° so ∠ EOC = 2x = 72 °
As shown in the figure, ∠ AOB = 35 ° 40 ′, ∠ BOC = 50 ° 30 ′, ∠ doc = 21 ° 18 ′, OE bisects ∠ AOD, and calculates the degree of ∠ BOE
According to the OE, divide the AOD, AOE = 12, AOD = 53 ° 44 ', and BOE = AOE - AOB = 53 ° 44' - 35 ° 40 '= 18 ° 4'
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