কোণের অন্তর্দ্বিখণ্ডকসমূহ অন্তঃকেন্দ্রে মিলিত হয়

The angle bisectors of $\Delta ABC$ concur at a point called the incenter $I$. The incircle is the circle inscribed in $\Delta ABC.$ This circle has its center at $I.$

PROOF:

We will first suppose that the angle bisectors of $\angle ABC$ and $\angle BCA$ concur at a point called the incenter. Let us drop the perpendicular from $I$ to the $3$ sides of the triangle. Since $AF$ and $AE$ are tangent to $IF$ and $IE$, $AF = AE$. And similarly $BF = BD.$ Using the AAS congruence theorem, we have $\Delta AFI \cong \Delta AEI$ and $\Delta BFI \cong \Delta BDI.$ $\Rightarrow FI = ID$ and $FI = IE.$ Hence, $ID = IE.$ Now, using the SAS congruence theorem, we find that $\Delta CEI \cong \Delta CDI.$ Hence, $\angle ECI = \angle DCI.$ We will now proceed that there is a unique incircle. We realize that $ID, IE,$ and $IF$ are the shortest distances to the sides of $\Delta ABC$ from $I.$ In addition, we proved earlier that they are all equal. We also know that any point equidistant from the sides of $\Delta ABC$ must lie on the angle bisectors of $\Delta ABC.$ Hence $I$ is the center of the incircle of $\Delta ABC.$

$\blacksquare$

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We will now state one very useful relation:

$$\boxed{[ABC] = rs}$$

where $[ABC]$ is the area of $\Delta ABC$, $r$ is the radius of the incircle of $\Delta ABC$, and $s$ is the semiperimeter of $\Delta ABC$.

PROOF:

Let the side opposite to $\angle A = a$, the side opposite to $\angle B = b$, and the side opposite to $\angle C = c.$

$$[ABC] = [BCI]+[ACI]+[ABI]$$ $$\Rightarrow [ABC] = \frac{ar}{2}+\frac{br}{2}+\frac{cr}{2}$$ $$\Rightarrow [ABC] = r\left(\frac{a+b+c}{2}\right)$$ $$\Rightarrow [ABC] = rs.$$

$\blacksquare$

CONCLUSION:

Incenters and Incircles are very common in high school olympiads. It is essential to know how they work.