wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

With usual notations, prove that in a triangle ABC
bcr1+car2+abr3=0

Open in App
Solution

In any given triangle ABC, with usual notations, we have a relation
r1=Δsa
r2=Δsb
r3=Δsc
where, Δ is Area of given triangle,
a,b,c are sides of triangle and s is semi- perimeter,
r1,r2,r3 are the radii of ex-circles opposite to the vertices of A, B ,C of triangle ABC
Now, putting above relation of r1, r2 and r3 in the left side of given equality, we get
bc(Δsa)+ca(Δsb)+ab(Δsc)=0

On rearranging, we get
(bc)(sa)Δ+(ca)(sb)Δ+(ab)(sc)Δ=0 =0
Now multiplying both side by Δ, equation reduces to
(bc)(sa)+(ca)(sb)+(ab)(sc)=0
on solving further by opening brackets,
bsbacs+ca+cscbas+ab+asacbs+bc=0
In left side of equation each term gets cancelled and we obtain
0=0
hence proved.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon