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In △ ABC, if the bisectors of ∠ a = 50 °, B and C intersect at point O, then the degree of ∠ BOC is () A. 65° B. 100° C. 115° D. 130°
∵ a = 50 °, the bisectors be and CF intersect at o,
∴∠OBC+∠OCB=1
2（∠ABC+∠ACB）=65°，
∴∠BOC=180°-65°=115°，
Therefore, C
It is known that in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o It is proved that: ∠ BOC = 90 ° + 1 2∠A．
It is proved that: the bisector of ABC and ∠ ACB intersects at the point O, OBC = 12 ∠ ABC, OCB = 12 ∠ ACB, ? OBC + ∠ OCB = 12 (∠ ABC + ∠ ACB). In △ OBC, ∠ BOC = 180 ° - (∠ OBC + ∠ OCB) = 180 ° - 12 (180 ° - ∠ a) = 90 ° + 12 ∠ a
As shown in the figure, if ⊙ o cuts the three sides of ⊙ ABC, the chord length obtained by ⊙ o cutting the three sides of ⊙ ABC is equal, then the degree of ﹤ BOC is () A. 160° B. 135° C. 125° D. 110°
∵ a = 70 ° in ∵ ABC, ⊙ o cuts the three sides of ⊙ ABC, and the chord length is equal,
The distance from O to the three sides of the triangle is equal, that is, O is the heart of △ ABC,
∴∠1=∠2，∠3=∠4，∠1+∠3=1
2（180°-∠A）=1
2（180°-70°）=55°，
∴∠BOC=180°-（∠1+∠3）=180°-55°=125°．
Therefore, C
As shown in the figure, ⊙ o has the same chord length as ⊙ a = 70 ° on the three sides of ⊙ ABC
O is used to make om ⊥ AB, on ⊥ AC, Op ⊥ BC, and the vertical feet are m, N, P, respectively,
∵DE=FG=HI
∴OM=OP=ON
ν o is the intersection point of ∠ B, ∠ C bisectors
∵∠A=70°，
∴∠B+∠C=180°-∠A=110°，
And ∵ o is the intersection of ∵ B, ∵ C bisector,
∴∠BOC=180°-1
2（∠B+∠C）=180°-1
2×110°=125°．
As shown in the figure, if ⊙ o cuts the three sides of ⊙ ABC, the chord length obtained by ⊙ o cutting the three sides of ⊙ ABC is equal, then the degree of ﹤ BOC is () A. 160° B. 135° C. 125° D. 110°
∵ a = 70 ° in ∵ ABC, ⊙ o cuts the three sides of ⊙ ABC, and the chord length is equal,
The distance from O to the three sides of the triangle is equal, that is, O is the heart of △ ABC,
∴∠1=∠2，∠3=∠4，∠1+∠3=1
2（180°-∠A）=1
2（180°-70°）=55°，
∴∠BOC=180°-（∠1+∠3）=180°-55°=125°．
Therefore, C
In triangle ABC, if angle a equals 70 degrees and circle O cuts the three sides of triangle ABC with equal chord length, then what degree is BOC equal to?
It can be seen that OB bisector angle B, OC bisector angle C
Angle BOC = 180 - angle OBC - angle OCB = 180 - (angle B + angle c) / 2 = 180 - (180 - angle a) / 2 = 180-110 / 2 = 125 degrees
Point O is the outer center of triangle ABC, point I is the inner center of triangle ABC, and angle BIC = angle BOC, calculate the degree of angle A
Because ∠ BIC = 90 + 1 / 2 ∠ a, ∠ BOC = 2 ∠ a
So 90 + 1 / 2 ∠ a = 2 ∠ a
So 180 = 3 ∠ a
So ∠ a = 60 degrees
O in △ ABC, OA (vector) + 2ob (vector) + 3oc (vector) = 0, calculate the ratio of △ ABC and △ AOC area There is a general process, the process is not very important, only as the basis for the correct results. But be sure to give! Correct! Results, thank you
S△ AOC:S Delta AOB:S △BOC=2:3:1
S△ ABC:S △AOC=3:1
Extend OB to B 'to make ob' = 2ob; extend OC to C 'to make OC' = 3oc;
Connect b'c ', take the midpoint D of b'c', connect od and extend to a ', so that Da' = OD;
A 'B' is a parallelogram
So 2ob + 3oc = ob '+ OC' = OA '
Because OA + 2ob + 3oc = 0
OA + OA '= 0, or AO = OA'
So a, O, a 'are collinear, and modulus Ao = module OA'
By using the equal area of the triangle with the same base and equal height
S△AOC=S△A'OC=S△OCB'=2S△BOC
S△AOB=S△A'OB=S△OBC'=3S△BOC
So s △ AOC:S Delta AOB:S △BOC=2S△ BOC:3S Delta BOC:S △BOC=2:3:1
That is, s (ABC): s (AOC) = 3:1
Let o be in the interior of △ ABC with OA+2 OB+3 OC＝ Then the ratio of the area of △ ABC to the area of △ AOC is 0___ .
Take the midpoint D and e of AC and BC respectively,
A kind of
OA+2
OB+3
OC=
0，
Qi
OA+
OC=-2(
OB+
OC), which is 2
OD=-4
OE，
/ / O is a trisection of de,
∴S△ABC
S△AOC=3，
So the answer is: 3
It is known that in △ ABC, if the angular bisectors of ∠ a = 80 °, ABC and ACB intersect at point O, then ∠ BOC=______ .
As shown in the figure:
∵∠A=80°，
∴∠ABC+∠ACB=100°，
∵ point O is the intersection point of the angular bisector of ᙽ ABC and ᙽ ACB,
∴∠OBC+∠OCB=50°，
∴∠BOC=130°．
So the answer is: 130 degrees
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