Question

in $\triangle abc$, $cj = 18$. if $cg = bg$, what is $kj$?
a. 3
b. 6
c. 9
d. 12

Explanation:

Step1: Identify Medians and Midsegments

In $\triangle ABC$, since $CG = BG$, $G$ is the midpoint of $BC$. Also, the markings on $AC$ and $AB$ suggest $I$ and $J$ are midpoints (so $AI = IC$ and $AJ = JB$), making $IJ$ a midsegment, and $K$ (intersection of medians) the centroid? Wait, no—wait, $CJ = 18$, and we need $KJ$. Wait, first, since $CG = BG$, $AG$ is a median? Wait, no, the key is: in a triangle, the centroid divides a median into a ratio of $2:1$ (from vertex to midpoint). But also, if $I$ and $J$ are midpoints, $IJ \parallel BC$ and $IJ = \frac{1}{2}BC$. But wait, the problem: $CJ = 18$, and we need $KJ$. Wait, maybe $K$ is the centroid? Wait, no, let's re-examine. Wait, the diagram: $H$ is on $BC$ with $CH \perp AH$? No, $H$ has a right angle, but maybe $G$ is midpoint, so $AG$ is a median, and $CJ$ is a median? Wait, $CJ$: since $J$ is midpoint of $AB$ (marking on $AB$: one tick, so $AJ = JB$), so $CJ$ is a median. Then, the intersection of medians is the centroid, which divides each median into $2:1$. Wait, but also, $I$ is midpoint of $AC$ (two ticks, $AI = IC$), so $IJ$ is midline, parallel to $BC$, length $\frac{1}{2}BC$. But the question is $KJ$. Wait, maybe $K$ is the midpoint of $CJ$? No, wait, the options: 3,6,9,12. $CJ = 18$. If $K$ is the centroid, then $CK:KJ = 2:1$, so $KJ = \frac{1}{3}CJ$? Wait, no, centroid divides median into $2:1$ (from vertex to midpoint). Wait, $CJ$ is a median (since $J$ is midpoint of $AB$), so the centroid would be on $CJ$? Wait, but also, $I$ is midpoint of $AC$, so $AI = IC$, and $J$ is midpoint of $AB$, so $AJ = JB$. Then $IJ$ is midline, so $IJ \parallel BC$ and $IJ = \frac{1}{2}BC$. Then, the line from $C$ to $J$ (median) and the line from $A$ to $G$ (median, since $G$ is midpoint of $BC$) intersect at centroid $K$. Then, centroid divides each median into $2:1$, so $CK = \frac{2}{3}CJ$ and $KJ = \frac{1}{3}CJ$? Wait, no, wait: median length from $C$ to $J$ (midpoint of $AB$) is $CJ = 18$. Then centroid $K$ is on $CJ$, so $CK:KJ = 2:1$. Thus, $KJ = \frac{1}{3} \times 18 = 6$? Wait, no, wait: $2:1$ ratio, so total parts 3. $KJ$ is 1 part, so $18 \times \frac{1}{3} = 6$? Wait, but the options have 6 as B. Wait, but maybe I made a mistake. Wait, another approach: $IJ$ is midline, so $IJ \parallel BC$, so triangle $IKJ$ similar to triangle $CKG$? No, maybe simpler: $J$ is midpoint of $AB$, $I$ is midpoint of $AC$, so $IJ$ is midline, so $IJ \parallel BC$ and $IJ = \frac{1}{2}BC$. Then, $K$ is the midpoint of $CJ$? No, $CJ = 18$, if $K$ is midpoint, $KJ = 9$, but that's option C. Wait, confusion. Wait, let's check the markings: $AC$ has two ticks (so $AI = IC$), $AB$ has one tick (so $AJ = JB$), $BC$: $CG = BG$ (so $G$ is midpoint). So $IJ$ is midline (connecting midpoints of $AC$ and $AB$), so $IJ \parallel BC$, $IJ = \frac{1}{2}BC$. Then, the line $CJ$: $J$ is midpoint of $AB$, so $CJ$ is median. The intersection of $IJ$ and $AG$ (median from $A$ to $G$) is $K$. Since $IJ \parallel BC$, triangle $IKJ \sim$ triangle $CKG$ (by AA similarity, since $IJ \parallel BC$, angles equal). The ratio of similarity: $IJ:BC = 1:2$, so $IK:KC = 1:2$, and $JK:KG = 1:2$. But $CJ$ is a median, length 18. Wait, maybe $K$ is the midpoint of $IJ$? No, this is getting confusing. Wait, the answer options: 3,6,9,12. $CJ = 18$. If $KJ = 6$, that's $\frac{18}{3} = 6$ (if centroid, 2:1 ratio, so $KJ = 6$). Wait, centroid divides median into $2:1$, so from $C$ to $J$, centroid $K$: $CK = 12$, $KJ = 6$. Yes, that makes sense. So $KJ = \frac{1}{3}CJ = \frac{18}{3} = 6$.

Step2: Calculate $KJ$

Since $K$ is the centroid (intersection of medians), it divides the median $CJ$ into a ratio of $2:1$ (from vertex $C$ to midpoint $J$). Thus, $KJ = \frac{1}{3} \times CJ$.
Given $CJ = 18$,
$KJ = \frac{1}{3} \times 18 = 6$.

Answer:

$\boxed{6}$ (Option B)