Given the problem, we have a right-angled isosceles triangle with angles $\alpha = 45^\circ$, $\beta = 45^\circ$, and $\gamma = 90^\circ$. The sides opposite these angles are $a = 4 \, \text{cm}$, $b = 4 \, \text{cm}$, and the hypotenuse $c = \sqrt{a^2 + b^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \, \text{cm}$.

When the angle bisectors of a triangle are constructed, they meet at the incenter of the triangle. The angles of the new triangle formed by the intersection points of the angle bisectors, typically denoted as triangle $KLM$, are half the sum of the angles of the original triangle.

Calculating the angles of triangle $KLM$:

1. Angle at $K$ (opposite $\alpha = 45^\circ$):

   • The angle at $K$ will be half the sum of the other two angles $\beta + \gamma$: $\angle K = \frac{1}{2} \times (45^\circ + 90^\circ) = \frac{135^\circ}{2} = 67.5^\circ$

2. Angle at $L$ (opposite $\beta = 45^\circ$):

   • The angle at $L$ will be half the sum of the other two angles $\alpha + \gamma$: $\angle L = \frac{1}{2} \times (45^\circ + 90^\circ) = \frac{135^\circ}{2} = 67.5^\circ$

3. Angle at $M$ (opposite $\gamma = 90^\circ$):

   • The angle at $M$ will be half the sum of the other two angles $\alpha + \beta$: $\angle M = \frac{1}{2} \times (45^\circ + 45^\circ) = \frac{90^\circ}{2} = 45^\circ$

Therefore, the angles of the triangle $KLM$ are:

• $\angle K = 67.5^\circ$
• $\angle L = 67.5^\circ$
• $\angle M = 45^\circ$

These angles add up to $180^\circ$, confirming the calculation.

Would you like more details on this topic, or do you have any other questions?

Related Questions:

1. What is the significance of the incenter in a triangle?
2. How do you calculate the length of the incenter in an isosceles right triangle?
3. What is the relationship between the circumcenter and the incenter in a right triangle?
4. How do the angles of a triangle affect the properties of its angle bisectors?
5. How do you construct the triangle $KLM$ geometrically?

Tip:

In any triangle, the angle bisectors always intersect at the incenter, which is equidistant from all the sides of the triangle.