As shown in the figure, the straight lines AB and CD intersect at the points o, OE and of, which are bisectors of ∠ AOC and ∠ AOD respectively. It is known that ∠ BOC = 70 ° to find the degree of ∠ DOF and ∠ Coe

As shown in the figure, the straight lines AB and CD intersect at the points o, OE and of, which are bisectors of ∠ AOC and ∠ AOD respectively. It is known that ∠ BOC = 70 ° to find the degree of ∠ DOF and ∠ Coe

Angle cob plus angle BOF plus angle FOD equals 180 degrees, AOE plus angle FOC plus angle cob equals 180 degrees, double angle equals 180 minus 70 equals 110. Double angle equals 180 minus 70 equals 110. So angle EOC equals angle DOF equals 55 degrees

As shown in the figure, point O is on the straight line AB, the ratio of degree between ∠ AOE and ∠ BOC is 5:3, OD bisects ∠ Coe, ∠ AOC = 3 ∠ AOE, and calculates the degree between ∠ AOC and ∠ BOD

The ratio of degree between AOE and BOC is 5:3
Let BOC = 3x and AOE = 5x
Because ∠ AOC = 3 ∠ AOE
.∠AOC=15x,∠COE=10x
Od bisection ∠ Coe
∠EOD=∠COD==1/2coe=5x
aoc+cob=180
3x+15x=180
x=10
∠AOC=150° ∠BOD=80°

As shown in the figure, O is a point on the straight line AB, OC is any ray, OD bisecting angle BOC, OE bisecting angle AOC, try to explain the quantitative relationship between angle COD and angle Coe
It is helpful for the responder to give an accurate answer

Angle cod + angle COE = 90 degrees

As shown in the figure, OE and of are bisectors of AOC and BOC respectively, and OE ⊥ of

It is proved that: O E and of are bisectors of AOC and BOC, AOE = Coe, BOF = COF, OE ⊥ of, COE + COF = 90 degree, BOF = 90 degree, AOB = COE + COF + BOF = 90 degree, AOE + BOF = 180 degree, BOF = 90 degree, AOA, O and B are on the same line