Given angles A,B,C are in AP and A+B+C=180∘‌‌(∵ angles of △ABC) So, ‌
A+C
2
=B ‌∴2B+B=180∘ ‌⇒‌‌3B=180∘ ‌⇒‌‌B=60∘ Given that r32=r1r2+r2r3+r3r1 Since, B=60∘, using the cosine rule for side b. b2‌=a2+c2−2ac‌cos‌60∘ ‌=a2+c2−2ac⋅‌
1
2
=a2+c2−ac We know that, s2=r1r2+r2r3+r3r1 And given, r32=r1r2+r2r3+r3r1 So, r32=s2 ⇒r3=s Since, r3=‌
∆
s−c
‌⇒‌‌s=‌
∆
s−c
⇒∆=s(s−c) ‌⇒‌‌√s(s−a)(s−b)(s−c)=s(s−c) Squaring both sides, we get ‌⇒s(s−a)(s−b)(s−c)=s2(s−c)2 ‌⇒(s−a)(s−b)=s(s−c) ‌⇒s2−sa−sb+ab=s2−sc ‌⇒ab=s(a+b−c) ‌⇒ab=‌
(a+b+c)
2
⋅(a+b−c) ‌‌‌[∵s=‌
a+b+c
2
] ‌⇒2ab=(a+b)2−c2 ‌⇒2ab=a2+b2+2ab−c2 ‌⇒a2+b2−c2=0 ‌⇒c2=a2+b2 So, the triangle is a right-angled triangle with right angle at c. But, we have B=60∘ and C=90∘ So, A=180∘−60∘−90∘=30∘ Using the sine rule,‌‌