In triangle ABC, angle a is equal to angle abd, angle ABC is equal to angle c, angle c is angle BDC, and degree of angle DBC is calculated

In triangle ABC, angle a is equal to angle abd, angle ABC is equal to angle c, angle c is angle BDC, and degree of angle DBC is calculated

fifty-five thousand five hundred and fifty-five

As shown in the figure, BD bisection angle ABC, CD bisection angle ace, given the degree of angle a is 80, calculate the degree of angle BDC

Prolonged BD crossed AC to g
∵∠A＝80
∴∠ABC+∠ACB＝180-∠A＝180-80＝100
∵ BD bisection ∠ ABC, CD bisection  ACB
∴∠ABG＝∠ABC/2,∠ACD＝∠ACB/2
∴∠BGC＝∠A+∠ABG＝∠A+∠ABC/2
∴∠BDC＝∠BGC+∠ACD＝∠A+∠ABC/2+∠ACB/2＝∠A+（∠ABC+∠ACB）/2＝80+50＝130°
The math group answered your question,

As shown in the figure, D is any point in △ ABC, and it is proved that: ∠ BDC ＞∠ a (the reasoning basis can be omitted in the process of proof.)

It was proved that the extended BD was crossed with AC to E
∵ BDC is an external angle of  Dec,
∴∠BDC＞∠DEC，
And ∵ Dec is an external angle of ∵ Abe,
∴∠DEC＞∠A，
∴∠BDC＞∠A．

As shown in the figure, D is any point in △ ABC, and it is proved that: ∠ BDC ＞∠ a (the reasoning basis can be omitted in the process of proof.)

It was proved that the extended BD was crossed with AC to E
∵ BDC is an external angle of  Dec,
∴∠BDC＞∠DEC，
And ∵ Dec is an external angle of ∵ Abe,
∴∠DEC＞∠A，
∴∠BDC＞∠A．

As shown in the figure, the triangle ABC is an equilateral triangle, and the angle BDC is 120 Confirmation: ad = BD + CD

DC, de = DC
Since Δ ABC is an equilateral triangle, then ∠ a = 60 °, therefore ∠ a + ∠ BDC = 120 °
So points a, B, D and C are in a circle
So we have ∠ ADC = ∠ ABC = 60 ° so Δ CDE is a regular triangle
Thus, there is ∠ DCB = ∠ ECA = 60 ° - ∠ BCE, and DC = EC
And BC = AC
So Δ DCB ≌ Δ ECA, then BD = AE can be obtained
Therefore, ad = AE + be = BD + CD

It is known that: as shown in the figure, in △ ABC, the bisector of ∠ B and the bisector of external angle of △ ABC intersect at point D, ∠ a = 90 °

∵ BD bisection ∵ ABC,
∴∠CBD=1
2∠ABC，
∵ CD bisects ᙽ the outer angle of ABC,
∴∠DCE=1
2∠ACE=1
2（∠A+∠ABC）=1
2∠A+1
2∠ABC，
In △ BCD, by the exterior angle property of triangle, ∠ DCE = ∠ CBD + ∠ d = 1
2∠ABC+∠D，
∴1
2∠A+1
2∠ABC=1
2∠ABC+∠D，
∴∠D=1
2∠BAC=1
2×90°=45°．

As shown in the figure, in △ ABC, the bisector of [ABC] intersects the bisector of [ACB] at point E, and the extension line of [EC] intersects the bisector of [ABC] at point D. If [D] is 10 ° greater than [E], the degree of [a] is______ .

 be bisection ∠ ABC,  EBC = 12 ∠ ABC,  ABC = 2 ∠ EBC, ? BD bisection ∠ outside angle of ABC, ? CBD = 12 ∠ CBF, ∵ EBC + ∠ CBD = 12 ∠ ABC + 12 ∠ CBF = 12 (∠ ABC + ∠ CBF) = 12 ∠ ABF = 12 × 180 ° = 90 °, i.e., ∠ EBD = 90 °, ? D + ∠ e = 90 °, ? D - ∠ e = 10 °

As shown in the figure, D is the intersection of the bisector of angle ABC in triangle ABC and the bisector of angle ACB. It is proved that angle a is equal to angle D

Here, just use the theorem that "an outer angle of a triangle is equal to the sum of two nonadjacent interior angles"
Take point E on BC extension line
Then ∠ a = ∠ ace - ∠ ABC, ∠ d = ∠ DCE - ∠ DBC
Because ∠ ace = 2 ∠ DCE, ∠ ABC = 2 ∠ DBC
So ∠ a = 2 ∠ D

As shown in the figure, D is a point in △ ABC, ∠ abd = 20 °, ACD = 25 °, a = 35 ° to find the degree of ∠ BDC

∵∠A+∠ABC+∠ACB=180°
∴∠ABC+∠ACB=180°-35°=145°
∵∠ABC=∠ABD+∠DBC
∠ACB=∠ACD+∠DCB
∴∠ABD+∠DBC+∠ACD+∠DBC=∠ABC+∠ACB=145°
That is ∠ DBC + ∠ DCB = 145 ° - ∠ abd - ∠ ACD = 145 ° - 20 ° - 25 ° = 100 °
∴∠BCD=180°-(∠DBC+∠DCB)=180°-100°=80°

As shown in the figure, it is known that the bisectors of the exterior angles of △ ABC and ∠ ACB intersect at D, ∠ a = 40 ° and calculate the degree of ∠ BDC

 BD and CD are bisectors of  ABC and  ACB,  CBD = 12 (﹤ a + ∠ ACB), ∠ BCD = 12 (﹤ a + ∠ ABC), ? ABC + ∠ ACB = 180 ° - ∠ a, ? BDC = 180 ° - ∠ CBD - ∠ BCD = 180 ° - 12 (∠ a + ∠ ACB + ∠ a +) = 90 ° - 12