As shown in the figure, ab ∥ CD, ∠ bed = 110 °, BF bisection ∥ Abe, DF bisection ∥ CDE, then ∥ BFD = () A. 110°B. 115°C. 125°D. 130°

As shown in the figure, ab ∥ CD, ∠ bed = 110 °, BF bisection ∥ Abe, DF bisection ∥ CDE, then ∥ BFD = () A. 110°B. 115°C. 125°D. 130°

The over point F is FN \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\= 125 °, ∵ ∠ DFN = ∠ CDF, ∠ BFN= Therefore, C is selected

As shown in the figure, if AB is parallel to CD, BF bisector angle Abe, DF bisector angle CDE
(1) (2) if ∠ bed = 75 °, calculate the degree of ∠ BFD

(1) We prove that if we make a straight line eg parallel to ab through e, then CD ∥ ab ∥ eg gets ∥ Abe = & nbsp;, ∥ CDE = & nbsp; ∥ DEG ∵ & nbsp; ∥ bed = & nbsp; ∥ beg + & nbsp; ∥ DEG & nbsp; ∥ & nbsp; ∥ Abe + ∥ CDE = ∥ bed (2), then we need to consider the quadrilateral fdeb

In the triangle ABC, the bisector of ∠ a intersects BC at point D, ∠ B = 65 ° and ∠ C = 50 °. The degree of ball ∠ ADB

∠A=180°-(∠B+∠C)=180°-(65°+50°)=65°
And ∵ BC is the bisector of ∠ a
∴∠BAD=65°÷2=32.5°
∴∠ADB=180°-(∠B+∠BAD)=180°-(65°+32.5°)=82.5°

As shown in the figure, in the isosceles triangle △ ABC, ∠ ABC = ∠ C, BD is the bisector of ∠ ABC, ∠ abd = 35 ° to find the degree of ∠ ADB

∵ BD is the bisector of ∠ ABC
∴∠CBD=∠ABD=35,∠ABC=2∠ABD=70°
∵∠ADB=∠CBD+∠C
∴∠ADB=105