As shown in the figure, △ ABC, ab = AC, BC = 8, BD is the middle line on the side of AC, and the difference between the circumference of △ abd and △ BDC is 2, then ab=______ .

As shown in the figure, △ ABC, ab = AC, BC = 8, BD is the middle line on the side of AC, and the difference between the circumference of △ abd and △ BDC is 2, then ab=______ .

∵ BD is the center line on the edge of AC, ∵ ad = CD, ∵ abd and △ BDC circumference difference = (AB + AD + BD) - (BC + CD + BD) = ab-bc, ∵ abd and △ BDC circumference difference is 2, BC = 8, ∵ AB-8 = 2, ∵ ab = 10
In △ ABC, ab = AC, BD is the middle line of △ ABC. It is known that the perimeter difference between △ abd and △ BDC is 6, and the perimeter of △ ABC is 30
AB = AC = 12, BC = 6 or AB = AC = 8, BC = 14
As shown in the figure, in triangle ABC, be and CD are bisectors of angles, and the intersection point is o. when ∠ a = 60 degree, (1), calculate the degree of ∠ BDC;
As shown in the figure, in triangle ABC, be and CD are bisectors of angles, and the intersection point is o. when ∠ a = 60 degree,
(1) Calculate the degree of ∠ BDC;
(2) When ∠ a = 100 °, calculate the degree of ∠ BOC;
(3) If ∠ a = α °, calculate the degree of ∠ BDC
∵∠A+∠ABC+∠ACB＝180
∴∠ABC+∠ACB＝180-∠A
∵ be bisection ∠ ABC
∴∠CBE＝∠ABC/2
∵ CD bisection ∠ ACB
∴∠BCD＝∠ACB/2
∴∠BOC＝180-（∠CBE+∠BCD）
＝180-（∠ABC/2+∠ACB/2）
＝180-（∠ABC+∠ACB）/2
＝180-（180-∠A）/2
＝90+∠A/2
Qi
(1) When ∠ a = 60, ∠ BOC = 90 + 60 / 2 = 120
(2) When ∠ a = 100, ∠ BOC = 90 + 100 / 2 = 140
(3) When ∠ a = α, ∠ BOC = 90 + α / 2
As shown in the figure, be and CF are the angular bisectors of △ ABC, ∠ a = 65 °, then BDC equals ()
A. 122.5°B. 187.5°C. 178.5°D. 115°
Therefore, a
Finding definite integral ∫ sinxcos ^ 3x / (1 + cos ^ 2x) DX
Let sinxdx = - D (cosx)
t^3/(1+t^2)dt
= [(t^3+t)-t]/(1+t^2) *dt = t-t/(1+t^2)
t^2/2 - 1/2 * ln(1+t^2) +C
cosx^2/2 - ln(1+cosx^2)/2 + C
In the right isosceles triangle ABC, AC = BC, and a point P in the triangle ABC satisfies PA = 3, PC = 2, Pb = 1, find the degree of angle BPC?
A new triangle CBP 'is obtained by rotating the triangle ACP around point C so that Ca and CB coincide. Then there is ∠ BCP' = ∠ ACP, so ∠ PCP '= ∠ PCB + ∠ BCP' = ∠ PCB + ∠ ACP = 90 ° and CP = CP '= 2. According to Pythagorean theorem, PP' = 2 √ 2 ∠ P ` PC = 45 ° and BP '= AP = 3PB = 13 * 3 = 2 √ 2 * 2 √ 2 + 1 * 1, so ∠ P ` Pb = 9
How to calculate ∫ [Tan (3x + 1) / cos ^ 2 (3x + 1)] DX,
∫[tan(3x+1)/cos^2(3x+1)]dx
=∫tan(3x+1)sec^2(3x+1)dx
=∫tan(3x+1)dtan(3x+1)
=(1/2)tan^2(3x+1)+c.
In p-abc, PA = Pb = PC = BC and BAC = 90, what is the angle between PC and ABC
Because PA = Pb = PC, and ∠ BAC is a right angle, because the projection of the vertex of a triangular pyramid on the bottom is the intersection of the median lines of the three sides of the triangle, that is, the point with equal distance to the three vertices, and because △ ABC is a right triangle, the projection of the vertex P of a triangular pyramid on the bottom is the midpoint of the line BC, and △ PBC is an equilateral triangle
Very urgent ∫ DX / cos & # 178; 3x
The original formula = (1 / 3) ∫ D (3x) / cos & # 178; 3x = (1 / 3) Tan (3x) + C
If a point P in the acute angle △ ABC satisfies PA = Pb = PC, then p is △ ABC
A. The intersection of three bisectors
B the intersection of the three midlines
The intersection of three heights of C
D the intersection of the vertical bisectors of three sides
D
Choose D. Because PA = Pb, P is on the vertical bisector of ab;
Because PA = PC, P is on the vertical bisector of AC
Because PC = Pb, P is on the vertical bisector of BC
So p is the intersection of the vertical bisectors of the three sides.
D
Choice: D
Reasons: 1. The distance from the point on the vertical bisector of the line segment to both ends of the line segment is equal
2. Equivalent substitution
This problem needs the above two knowledge points to solve.
D
The distance from one point to both ends of the bisector is equal
The intersection of the vertical bisectors of three sides D should be selected
Because PA = Pb, P is on the vertical bisector of ab;
Because PA = PC, P is on the vertical bisector of AC
Because PC = Pb, P is on the vertical bisector of BC
So p is the intersection of the vertical bisectors of the three sides
D