# Triangle side relation with incircle

Lets say I have to draw an incircle of radius R in a triangle with side lengths a,b and c.

Can I say that no side of all the possible triangles that can contain the in circle of radius R will be less than the length of R ?

For example,

I have to draw an in circle of radius 5 units. Can I say that all the triangles that can contain the in circle of Radius 5 will have no side less than 5 units?

actually no side can be less than or equal to $Diameter = 2*R$

when $Diameter = 2*R$ the limiting "2 infinite sides isosceles triangle" case gives the 2 sides infinite in length and parallel with base interior angles of $\frac{\pi}{2}$

just seeing the limiting case, applying geometric intuition to other cases seems to be "intuitive proof"

• This could be improved by a proof that each side is greater than $2R.$ Or maybe just a link to a proof... – coffeemath Jan 16 '17 at 5:44
• @coffeemath Is there such a proof available ? – ng.newbie Jan 16 '17 at 6:01

For any $\triangle ABC$, the centre of the incircle $I$ is at the intersection of the angle bisectors.

Therefore in $\triangle ABI$ (and similarly in $\triangle BCI$ and $\triangle CAI$), we know that $\angle AIB$ must be obtuse, since the original angles at $A$ and $B$ total to less than $180°$ therefore the half-angles $\angle ABI$ and $\angle IAB$ total to less than $90°$.

Thus the altitude from $I$ to $AB$ - which is the radius $R$ of the incircle - is less than half of $|AB|$. To see why this is true, draw a circle with $AB$ as the diameter and note that $I$ must be inside the circle (and thus, less than the radius of that $AB$-based circle from the diameter $AB$).

Therefore the radius of the incircle of any triangle is less than half the length of any of its sides.

Draw line $L$ through the incenter $I$ and parallel to side $AB.$ This line cuts through sides $AC,BC$ at some points $M,N.$

Now the incircle is interior to triangle $ABC,$ and so segment $MN$ contains a diameter of the incircle somewhere in it, showing the diameter of the incircle to be at most $MN.$ However one sees that $MN<AB$ since it is parallel to side $AB$ and goes through the point $I$ which is inside triangle $ABC.$

We can concude then that the diameter $2R$ of the incircle is indeed less than side $AB$ as desired. [Note the usual notation for the incircle radius is the lower case $r,$ upper case being used for circumcircle radius.]