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

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

Given the triangle ABC, ab = 7, BC = 8, AC = 5, its inscribed circle and ab are tangent to point D, then what is the length of ad?

Suppose that the inscribed circle and AB, BC, CA are tangent to points D, e, f respectively;
AD=X,
According to the tangent length theorem:
AF=AD=X,-------------------1
BE=BD=AB-AD=7-X
CF=CE=BC-BE=8-(7-X)=X+1
AF=AC-CF=5-(X+1)=4-X-------2
From 1,2, it is concluded that: 1
X=4-X
X=2=AD

In △ ABC, ∠ C = 90 °, BC = 3, AC = 4, its inscribed circle OI is tangent to the edge AB, BC, Ca at the points D, e, f respectively, and the value of ad times BD is calculated

∵BC=3,AC=4 ∴AB=5
The center of the inscribed circle of a triangle is the intersection of the bisectors of three angles
According to the meaning of the title
r=（a+b-c）/2=1
That is, the quadrilateral CEOF is a square
∴BE=3 AF=2
And