In the triangle ABC, if the bisector of the inner angle B and C intersects at point E and the bisector of the outer angle intersects at point F, and angle a is equal to 70 degrees, then angle BEC equals? Angle BFC equals?
The results show that: ABC + ACB = 180 degrees - a = 110 degrees; EB and EC are bisectors of angles, then: EBC + ECB = (1 / 2) ∠ ABC + (1 / 2) ∠ ACB = 55 degrees, so: ∠ BEC = 125 degrees. Be and BF are bisectors of inner and outer angles of ABC respectively, so be ⊥ BF can be easily obtained; similarly, EC ⊥ cf. then: FBC + ∠ FCB = (90 ° - ∠ EBC) + (90 ° -}
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- 1. It is known that in the triangle ABC, ∠ B = 1 / 2 ∠ a, CD is perpendicular to BC, CE is the middle line on the edge BD, and AC = 2 / 1ab is proved Using the properties of right triangle (1)
- 2. It is known that be is the bisector of the inner angle of triangle ABC and CE is the bisector of the outer angle of triangle ABC Add a reason after each step
- 3. As shown in the figure, in the triangle ABC, the angle a = 60 degrees. The bisectors BD and CE of the triangle ABC intersect at point F. if be = 4 and CD = 2, find the length of BC
- 4. In triangle ABC, bisectors be and CD of angle ABC and angle ACB intersect at point O. if angle a = 60 °, what is the relationship between BC, CE and BD
- 5. Two bisectors BD and CE of triangle ABC intersect at point O, angle a = 60 degrees, proving: CD + be = BC There is an answer to this question on the Internet, but I can't understand it because of my low IQ. Dashen explains why ∠ EOB = 1 / 2 (180 ° - 60 °) = 60 ° This is a certain method: take a point F on the edge of BC, make BF = be, link of ∵ BD is the angular bisector, BF = be, Bo is the common edge, ∴△BEO≌△BFO →∠EOB=∠FOB=∠COD ∵∠A=60° The angle bisector is BD and CE ∵∠EOB=1/2(180°-60°)=60° Then ∠ COF = 180 ° - ∠ FOB - ∠ cod = 60 ° And ∵ CE is the angular bisector and CO is the common edge, ∠ COF = ∠ cod = 60 ° (proved) ∴△COF≌△COD →CF=CD ∴BC=BF+CF=BE+CD Or there are other ways,
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- 15. As shown in the figure, △ ABC, point O is a moving point on the edge of AC. make a straight line Mn ‖ BC through point O. let the bisector of the intersection angle BCA of Mn be at point E, and the bisector of the external angle ACD of △ ABC be at point F (1) Try to explain the line segment EO = fo (2) Guess: when point O moves to where, the quadrilateral aecf is a rectangle? And explain the reason
- 16. In △ ABC, point O is a moving point on edge AC, and a straight line Mn parallel to BC is made through O. let Mn intersect the bisector of ∠ BCA at point e. the bisector of the outer angle of ∠ BCA intersects with point F 1. When point O moves on edge AC, will the quadrilateral BCFE be a diamond? If so, please prove. If not, explain the reason; 2. When point O moves to where, and △ ABC satisfies what conditions, the quadrilateral aecf is a square?
- 17. In △ ABC, point P is a moving point on edge AC, and a straight line Mn ∥ BC is made through point P. let Mn intersect ∥ BCA and its outer angle bisector be at point E, and intersect ∥ BCA and its outer angle be at point F (1) Verification: PE = PF (2) Can the quadrilateral BCFE be a diamond when point P moves on edge AC? Explain why (3) If there is a point P on the side of AC, let the quadrilateral aecf be a square, and AP / BC = (√ 3) / 2
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