Let's solve the problem systematically by addressing each question in order.

Given Information:

• $\triangle ABC$ is a right isosceles triangle:
  • $CB = CA = 3 \, \text{cm}$
  • $\angle C = 90^\circ$
• $C$ is the circle of diameter $[AB]$, with center $O$.
• $E$ is the symmetric point of $A$ with respect to $C$.

1)

(a) Calculate $AB$:

Using the Pythagorean theorem in $\triangle ABC$: $AB^2 = CB^2 + CA^2$ $AB^2 = 3^2 + 3^2 = 9 + 9 = 18 \implies AB = \sqrt{18} = 3\sqrt{2} \, \text{cm}.$

(b) Deduce the radius of the circle $C$:

The circle $C$ has $[AB]$ as its diameter. Hence: $\text{Radius} = \frac{\text{Diameter}}{2} = \frac{AB}{2} = \frac{3\sqrt{2}}{2} \, \text{cm}.$

2)

(a) Prove that $\triangle ABE$ is a right isosceles triangle:

• $E$ is the symmetric point of $A$ with respect to $C$, so:
  • $C$ is the midpoint of $[AE]$.
  • $AC = CE = 3 \, \text{cm}$.
• Since $CB = CA = 3 \, \text{cm}$, and $\angle C = 90^\circ$, $\triangle ABE$ inherits the symmetry, making it right isosceles at $B$.

(b) Deduce $BE$:

Using the Pythagorean theorem in $\triangle ABE$: $BE^2 = AB^2 + AE^2.$ Since $AE = 2 \times AC = 6 \, \text{cm}$, and $AB = 3\sqrt{2} \, \text{cm}$: $BE^2 = (3\sqrt{2})^2 + 6^2 = 18 + 36 = 54 \implies BE = \sqrt{54} = 3\sqrt{2} \, \text{cm}.$

3)

(a) What is the nature of $\triangle AKB$? Justify your answer:

• $K$ lies on the circle $C$, and $AB$ is the diameter of the circle.
• By the Thales theorem, $\triangle AKB$ is a right triangle at $K$.

(b) Show that $G$ is the centroid of $\triangle ABE$:

• The centroid of a triangle divides each median in a $2:1$ ratio.
• By construction, $G$ is the intersection of the medians, hence it is the centroid of $\triangle ABE$.

(c) Calculate $OE$, then deduce $OG$:

• $O$ is the center of the circle, and $OE$ is a radius: $OE = \text{Radius of the circle} = \frac{3\sqrt{2}}{2} \, \text{cm}.$
• $G$ divides $OE$ in a $2:1$ ratio (from $O$ to $E$): $OG = \frac{2}{3} \times OE = \frac{2}{3} \times \frac{3\sqrt{2}}{2} = \sqrt{2} \, \text{cm}.$

(d) Prove that $[AG]$ cuts $[BE]$ at its midpoint:

• $G$ is the centroid of $\triangle ABE$, meaning it divides all medians into a $2:1$ ratio.
• Hence, $[AG]$ divides $[BE]$ at its midpoint.

Final Answers:

1. (a) $AB = 3\sqrt{2} \, \text{cm}$, (b) Radius $= \frac{3\sqrt{2}}{2} \, \text{cm}$.
2. (a) $\triangle ABE$ is right isosceles, (b) $BE = 3\sqrt{2} \, \text{cm}$.
3. (a) $\triangle AKB$ is a right triangle, (b) $G$ is the centroid, (c) $OG = \sqrt{2} \, \text{cm}$, (d) $[AG]$ cuts $[BE]$ at its midpoint.

Would you like additional details for any of these steps?

Related Questions:

1. How do you calculate the coordinates of $G$ geometrically?
2. Why does $[AG]$ intersect $[BE]$ at the midpoint?
3. Can we verify $BE$ geometrically using a coordinate system?
4. What are the properties of the circle passing through $A, B, K$?
5. How does symmetry simplify the calculation of $AE$?

Tip:

Always draw a clear diagram with labeled points and medians when solving centroid-related problems!