In the following triangle \(▵ABC\),

\(AI\), \(BI\) and \(CI\) are the interior **angle bisectors** of the triangle.

They intersect at the point \(I\),
which is the **incenter** of \(△ABC\).

The incenter point of a triangle always lies inside the triangle,
and is equidistant from all the sides of that triangle.

The **incircle** of a triangle is a largest circle which
can be contained in that triangle.
The center of an incircle is known as the **incenter**,
while radius of the incircle is called the **inradius**.

An incircle is also known as the **inscribed circle**.

An angle bisector is a line segment that bisects an angle in two equal angles.

In the above, \(∠ABC = 80° \)

When it is bisected by the segment \(BD\),

produces two equal angles of \(40°\).

So, \(BD\) is the angle bisector of \(∠ABC \).

- ✓ Incenter is the point of intersection of the interior angle bisectors.
- ✓ It is the centre of the incircle of the triangle.
- ✓ An incenter is equidistant from all the sides of the triangle .
- ✓ An incenter point always lies
**inside the triangle.** - ✓ In an equilateral triangle, the centroid, the orthocenter,
- the circumcenter and the
**incenter**coincide at the same point.

\(AI\), \(BI\) and \(CI\) are the angle bisectors.

\(I\) is the **incenter** of the triangle.

The circle is the incircle of the triangle.

\(IX\), \(IY\) and \(IZ\) are the inradii.

\(\therefore IX=IY=IZ\)

In right-angled triangles \(▵AZI\) and \(▵AXI, \;IX=IZ\),

hypotenuse \(\;AI\) is common to both.

So, by **RHS** rule,\(\quad▵AZI ≅ ▵AXI\),

and by **CPCT** rule, \(\quad AX = AZ\)

Similarly,\(\quad▵BXI ≅ ▵BYI, \;BX=BY\)

and\(\quad▵CYI ≅ ▵CZI, \;CY=CZ\)

\(AI\), \(BI\) and \(CI\) are the angle bisectors.

\(I\) is the **incenter** of the triangle.

The circle is the incircle of the triangle.

\(IX\), \(IY\) and \(IZ\) are the inradii.

If coordinates of the vertices \(A\), \(B\) and \(C\) are
\( (-18, 9) \), \( (11, 9) \) and \( (-9, 33) \) respectively, and

\(c = \) length of \(AB\),

\(a = \) length of \(BC\) and

\(b = \) length of \(CA\),

then the coordinates of the incenter
\(I(x, y)\) can be calculated with the following formula:

Coordinates of \(I(x, y) \)

\[ =\left(\frac{ax_1+bx_2+cx_3}{a+b+c}\;, \frac{ay_1+by_2+cy_3}{a+b+c}\right) \]
Using distance formula \(\; \sqrt{(y_2-y_1)^2 + (x_2-x_1)^2} \)** ,**

first we calculate the length of the sides of the triangle.

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