Triangle ABC, angle bisector of angle a intersects with D, BD = 3, DC = 4, find the inscribed circle radius of triangle ABC

SACD=1/2AC*ADSin A/2=1/2CD*AD Sin ADC
SBCD=1/2BC*ADSin A/2=1/2BD*AD Sin ADB
Sin ADC=Sin ADB
AC：BC=CD：BD=4：3
Let ABC be right angle ABC, then BC = 7, AC = 5.6, BC = 4.2 (3:4:5 Pythagorean theorem)
r=(b+c-a)/2= 2.8

It is known that in △ ABC, ab = AC = 13, BC = 10, then the radius of the inscribed circle of △ ABC is ()
A. 103B. 125C. 2D. 3

Let the radius of inscribed circle be r, ∵ AB = AC = 13, BC = 10, ∵ BF = 5, ∵ AF = 12, then s △ ABC = 12 × 12 × 10 = 60; and ∵ s △ ABC = s △ OAC + s △ OBC + s △ OAC = 12rab + 12rac + 12rbc = 12R (13 + 13 + 10) = 60

In △ ABC, ab = 8, BC = 6, AC = 10 are known. Calculate △ ABC to get the radius of inscribed circle
Such as the title
Please tell me and pay attention to your tone, uncivilized people!

Radius of inscribed circle r = triangle area s / half of triangle perimeter P. using Helen's formula to find the area s = P * (P-A) * (P-B) * (P-C) under the root sign, r = 2

In △ ABC, ab = AC = 5. BC = 6, find the length of inscribed circle radius

Using formula to find radius
The radius of the inscribed circle is r = 2S △ C, where s is the area of the triangle and C is the perimeter of the triangle
Helen formula & nbsp; s = radical [P (P-A) (P-B) (P-C)] & nbsp; & nbsp; where p = 1 / 2 & nbsp; & nbsp; (a + B + C)
&So the area of triangle ABC = 12
Radius = 3 / 2 & nbsp;

Take the midpoint D of BC side and connect ad
∵AB=AC
The ABC is an isosceles triangle
∵ D is the midpoint of BC
Therefore, ad = 4
Then s △ ABC = 1 / 2 × ad × BC = 12, let the radius of inscribed circle be r and the center of inscribed circle be o
Then s △ ABC = s △ OAB + s △ OBC + s △ OAC = 1 / 2 × R × C (C is perimeter) = 1 / 2 × R × 16 = 8R = 12, then r = 3 / 2

In angle ABC, angle a = angle B = 2, angle c, then angle C=_
Another solution: in an isosceles triangle, if the base angle is x degree and the top angle is y degree, then the algebraic expression containing x is used to express y and Y is obtained=_ Using the algebraic expression containing y to express x, we get X=____ .

36
180-2x
(180-Y)/2

As shown in Figure 2, angle ABC is equal to angle a'b'c '

Properties of bisector of half angle a 'B' C '

In the triangle ABC, a = 30 ', C = 45', a = 2, find the area of the triangle ABC

According to the sine theorem, a / Sina = C / sinc
Therefore, the solution of 2 / sin30 = C / sin45 is C = 2 √ 2,
According to the formula, s = 1 / 2xaxcxsinb = 1 / 2x2x2 √ 2xsin105 = 1 + √ 3
Therefore, the area of triangle ABC is 1 + √ 3

It is proved by the counter argument that at most one angle of triangle ABC can be a right angle
I know, but there are still complete steps to draw a picture, write the known and verify

Suppose there are at least two right angles in a triangle
When there are two right angles, then the sum of the internal angles of the triangle is more than 180 degrees, which is contradictory to the sum of the internal angles of the triangle
When there are three right angles, then the sum of the internal angles of the triangle is more than 180 degrees, which is contradictory to the sum of the internal angles of the triangle
So there are at least two right angles in the triangle
So there can only be at most one right angle in the triangle ABC

To the contrary: in triangle ABC, at least two angles are acute
What is the assumption that "at least two corners are acute"

Suppose there are two angles greater than or equal to 90 degrees
So the sum of the three angles is greater than 180 degrees
Because the angle of the triangle is 180 degrees
So the original hypothesis does not hold
So in triangle ABC, at least two corners are acute

In △ ABC, we know that a + C = 2B, Tana * Tanc = 2 + √ 3, and find the angles a, B and C

Then B = 60, (Tana + Tanc) / (1-tanatanc) = Tan (a + C) = Tan (180 - b) = - tanb = - tan60 = - 3, so Tana + Tanc = - 3 (1-tanatanc) = - 3 (1-2-3) = 3 + 3 (1) and tanatanc = 2 + 3 (2), (1) - (2) get Tana + t

Triangle ABC, angle bisector of angle a intersects with D, BD = 3, DC = 4, find the inscribed circle radius of triangle ABC