1. **State the problem:** We are given two triangles, $\triangle ABK$ and $\triangle ACK$, with some side lengths expressed in terms of $x$. We need to determine if these triangles are congruent, justify it, and find the values of $x$, $KC$, $AB$, $AC$, $BC$, and then find $PK$. 2. **Given:** - $AB = 19$ - $AC = 19$ - $BK = 5x$ - $AC = 6x + 7$ - $CP = 4x$ - $AB = 9x - 14$ 3. **Check congruence of $\triangle ABK$ and $\triangle ACK$:** - Both triangles share side $AK$. - Given $AB = AC = 19$, so $AB = AC$. - If $BK = CK$, then by Side-Side-Side (SSS) or Side-Angle-Side (SAS) postulate, the triangles are congruent. 4. **Find $x$:** - From $AB = 9x - 14$ and $AB = 19$, set equation: $$9x - 14 = 19$$ - Solve for $x$: $$9x = 19 + 14$$ $$9x = 33$$ $$x = \frac{33}{9} = \frac{11}{3}$$ 5. **Find $AC$:** - Given $AC = 6x + 7$, substitute $x = \frac{11}{3}$: $$AC = 6 \times \frac{11}{3} + 7 = 2 \times 11 + 7 = 22 + 7 = 29$$ - But earlier, $AC = 19$ was given, so there is a contradiction. We must clarify which value to use. 6. **Resolve contradiction:** - Since $AB = 19$ and $AB = 9x - 14$, we trust the equation to find $x$. - Then $AC = 6x + 7 = 6 \times \frac{11}{3} + 7 = 29$, which conflicts with $AC = 19$. - Possibly, the problem means $AC$ is variable, so we accept $AC = 29$. 7. **Find $BK$:** - $BK = 5x = 5 \times \frac{11}{3} = \frac{55}{3} \approx 18.33$ 8. **Find $KC$:** - Since $BK$ and $KC$ are parts of segment $BC$, and $CP = 4x = 4 \times \frac{11}{3} = \frac{44}{3} \approx 14.67$, but $KC$ is not directly given. - Assuming $KC = CP = 4x = \frac{44}{3}$ (if $P$ lies on $KC$), then $KC = \frac{44}{3}$. 9. **Find $AB$, $AC$, and $BC$:** - $AB = 19$ (given) - $AC = 29$ (from calculation) - $BC = BK + KC = \frac{55}{3} + \frac{44}{3} = \frac{99}{3} = 33$ 10. **Find $PK$:** - If $P$ lies on $KC$ such that $CP = 4x = \frac{44}{3}$, then $PK = KC - CP = \frac{44}{3} - \frac{44}{3} = 0$. - This suggests $P$ coincides with $C$, so $PK = 0$. **Final answers:** - $x = \frac{11}{3}$ - $KC = \frac{44}{3}$ - $AB = 19$ - $AC = 29$ - $BC = 33$ - $PK = 0$ **Justification for congruence:** - $\triangle ABK$ and $\triangle ACK$ share side $AK$. - $AB \neq AC$ based on values, so triangles are not congruent by SSS or SAS. - If problem states $AB = AC = 19$, then triangles could be congruent by SAS if $BK = CK$. - Since $BK \neq CK$ ($\frac{55}{3} \neq \frac{44}{3}$), triangles are not congruent. Hence, $\triangle ABK \not\cong \triangle ACK$ based on given data.