Properties of Triangles

Given that △ABC is an isosceles triangle with base BC, we want to find r1.
The known formulas are:
‌r=‌

∆

S

‌r1=‌

∆

s−a

Where r is the inradius, r1 is the exradius opposite the side a,∆ is the area of the triangle, and S is the semi-perimeter.
Let's explore the product r⋅r1 :
r⋅r1=‌

∆2

s(s−a)

=‌

s(s−a)(s−b)(s−c)

s(s−a)

Since the triangle is isosceles with b=c :
r⋅r1=(s−b)(s−c)=(s−b)2
In the context of an isosceles triangle:
s=‌

a+b+c

2

Since b=c, we have:
s−b=‌

a+b+c

2

−b=‌

a+c−b

2

Thus:
(s−b)2=(‌

a+c−b

2

)2=‌

a2

4

Rewriting this in terms of the circumcircle radius R :
‌

a

sin‌A

=2R
Therefore:r⋅r1=‌

a2

4

=‌

4R2sin‌2A

4

=R2sin‌2A
Hence, for the isosceles triangle triangle‌A‌B‌C with base BC, we find:
r1=R2sin2A