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