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As shown in the figure, ∠ a = 35 °, C = 60 ° in △ ABC, BD bisection ∠ ABC, de ‖ BC intersecting AB with E, then ∠ BDE=______ ，∠BDC=______ .
In ∵ ABC, ∵ a = 35 °, C = 60 ° and BD bisection ∠ ABC,
∴∠ABD=∠CBD=1
2∠ABC=85°，
∵DE∥BC，
∴∠BDE=∠CBD=85°，∠ADE=∠C=60°，
∴∠EDC=120°，
Then ∠ BDC = ∠ EDC - ∠ BDE = 35 °
So the answer is: 85 degrees; 35 degrees
As shown in the figure, BD is the bisector of ∠ ABC, de ‖ CB, intersecting AB at point E, ∠ a = 45 ° and ∠ BDC = 60 ° to find the degree of each internal angle of △ BDE
∵∠A=45°，∠BDC=60°，
∴∠ABD=∠BDC-∠A=15°．
∵ BD is the angular bisector of ∵ ABC,
∴∠DBC=∠EBD=15°，
∵DE∥BC，
∴∠BDE=∠DBC=15°；
∴∠BED=180°-∠EBD-∠EDB=150°．
As shown in the figure, BD is the bisector of ∠ ABC, de ‖ CB, intersecting AB at point E, ∠ a = 45 ° and ∠ BDC = 60 ° to find the degree of each internal angle of △ BDE
∵∠A=45°，∠BDC=60°，
∴∠ABD=∠BDC-∠A=15°．
∵ BD is the angular bisector of ∵ ABC,
∴∠DBC=∠EBD=15°，
∵DE∥BC，
∴∠BDE=∠DBC=15°；
∴∠BED=180°-∠EBD-∠EDB=150°．
BD is the bisector of the angle of triangle ABC. De parallels BC and intersects AB at point E. angle a = 45 ° and angle BDC = 60 ° to find the degree of each inner angle of triangle BDE?
15°/15°/150°
If a = 45 ° BDC = 60 ° DBE = 15 ° then BDE = DBC = DBE = 15 ° DEB = 150 °
As shown in Figure 7, in the triangle ABC, the angle ABC is equal to the angle c, D is a point on the edge of AC, the angle a is equal to the angle ADB, the angle DBC is equal to 30 ° and the degree of angle BDC is calculated
Let the angle BDC be x degree
2（180°-X-30°)+180°-X=180°
360°-2X-60°+180°-X=180°
360°-60°-2X-X=0
300°-3X=0
-3X=-300
3X=300
X=100
In △ ABC, ∠ a = 1 2∠C=1 If 2 ∠ ABC, BD are angular bisectors, then ∠ a=______ ，∠BDC=______ .
Let ∠ a = x, then ∠ C = 2x, ∠ ABC = 2x,
∵∠A+∠ABC+∠C=180°，
∴x+2x+2x=180°，
∴x=36°，
∴∠A=36°，∠ABC=2×36°=72°，
And ∵ BD is the angular bisector,
∴∠ABD=1
2∠ABC=36°，
∴∠BDC=∠A+∠ABD=36°+36°=72°．
So the answer is 36 °, 72 °
In △ ABC, ∠ a = 1 2∠C=1 If 2 ∠ ABC, BD are angular bisectors, then ∠ a=______ ，∠BDC=______ .
Let ∠ a = x, then ∠ C = 2x, ∠ ABC = 2x,
∵∠A+∠ABC+∠C=180°，
∴x+2x+2x=180°，
∴x=36°，
∴∠A=36°，∠ABC=2×36°=72°，
And ∵ BD is the angular bisector,
∴∠ABD=1
2∠ABC=36°，
∴∠BDC=∠A+∠ABD=36°+36°=72°．
So the answer is 36 °, 72 °
In the triangle ABC, BD is the bisector of an angle. Angle a = half angle c = half angle ABC. Calculate the degree of angle A and angle BDC
According to the triangle inner angle sum is 180 degrees angle a + angle B + angle c = 180 degrees → 5 times angle a = 180 degrees → angle a = 36 degrees, angle ABC = 72 degrees, angle c = 72 degrees, angle DBC = 36 degrees, then angle BDC = 180 degrees angle c-angle DBC = 180-72-36 = 36 degrees
In △ ABC, ∠ a = 1 2∠C=1 If 2 ∠ ABC, BD are angular bisectors, then ∠ a=______ ，∠BDC=______ .
Let ∠ a = x, then ∠ C = 2x, ∠ ABC = 2x,
∵∠A+∠ABC+∠C=180°，
∴x+2x+2x=180°，
∴x=36°，
∴∠A=36°，∠ABC=2×36°=72°，
And ∵ BD is the angular bisector,
∴∠ABD=1
2∠ABC=36°，
∴∠BDC=∠A+∠ABD=36°+36°=72°．
So the answer is 36 °, 72 °
As shown in the figure, in △ ABC, ∠ a = 62 °, 1 = 20 ° and ∠ 2 = 35 °. Calculate the degree of ∠ BDC
∵ in ᙽ ABC, ∵ a = 62 °,
∴∠ABC+∠ACB=180°-62°=118°．
∵∠1=20°，∠2=35°，
∴∠DBC+∠DCB=∠ABC+∠ACB-∠1-∠2=118°-20°-35°=63°．
∴∠BDC=180°-（∠DBC+∠DCB）=180°-63°=117°．
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