9.3. In a right triangle $A B C$ on the hypotenuse $A B$, a point $E$ is chosen such that $A C=C E$. The angle bisectors $C L$ and $E K$ of triangle $B C E$ intersect at point $I$. It is known that triangle $I K C$ is isosceles. Find the ratio $C L: A B$.
Solution: Angle $B$ must be less than angle $A$, otherwise there will be no point $E$ on the hypotenuse $AB$ with the property $AC = CE$. Let $\angle B = \beta$. Let, moreover, $\angle ECB = \gamma$, and $\angle CEB = \delta$. Clearly, $\beta + \gamma + \delta = 180^{\circ}$. The equalities $\angle CAE = 90^{\circ} - \beta$ and $\angle CEA = 180^{\circ} - \delta$ hold. Since these angles are equal (due to the equality of sides $AC$ and $CE$ in triangle $ACE$), we have $\delta = 90^{\circ} + \beta$,

![](https://cdn.mathpix.com/cropped/2024_05_06_7ed0fc493c4e4d38e847g-2.jpg?height=386&width=825&top_left_y=1760&top_left_x=618)

$\mathrm{K}$ to the solution of problem 9.3

from which $2\beta + \gamma = 90^{\circ}$. From triangle $BEC$, the angles of triangle $CIK$ are easily found: $\angle ICK = \gamma / 2$, $\angle CKI = \delta / 2 + \beta$, $\angle KIC = \gamma / 2 + \delta / 2$. Among these angles, there are two equal ones. The equality $\angle ICK = \angle KIC$ is impossible, since $\delta > 0$. The equality $\angle ICK = \angle CKI$ leads to the equation

$$
\gamma / 2 = \delta / 2 + \beta, \quad \gamma = \delta + 2\beta = 90^{\circ} + 3\beta = \gamma + 5\beta
$$

a contradiction. The only remaining possibility is $\angle CIK = \angle CKI$, which is equivalent to $\beta = \gamma / 2$. Then triangle $LCB$ is isosceles, $CL = LB$, and by the property of a right triangle, point $L$ is the midpoint of its hypotenuse. Then $CL : AB = 0.5$.

Note that the angles of $\triangle ABC$ are uniquely determined: $\angle C = 90^{\circ}$, $\angle B = 22.5^{\circ}$, $\angle A = 67.5^{\circ}$. This means that the described construction exists.

Answer: $CL : AB = 0.5$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct justified answer | 7 points |
| No proof of the existence of the construction mentioned in the problem | points not reduced |
| Solution is correct, except that it is not justified which sides of triangle $IKC$ are equal | 5 points |
| Correct answer, supported by a correct example of a triangle, without proof of its uniqueness | 3 points |
| Correct justification of the impossibility of both equalities 1) $CI = IK$ and 2) $CK = IK$ | 2 points |
| Correct justification of the impossibility of only one equality (see previous point) | 1 point |
| Correct answer without justification | 0 points |