Question

$AX,BY$ and $CZ$ are the medians of $△ABC$ intersecting at $O$. If $CK$ is drawn parallel to $BY$ to meet $AX$ in $K$, then $AO$ is

• A. $\frac{1}{4}$ $AK$
• B. $\frac{1}{2}$ $AK$
• C. $\frac{1}{3}$ $AK$
• D. $\frac{2}{3}$ $AK$

Answer 1

The correct option is B $\frac{1}{2}$ $AK$

Given:$AX,BY,CZ$ are medians of$△ABC$;They intersect atO.

$CK||BY$

To find:$AO$

Solution:$BY||CK$

$In△ACK$

$OY||KC$ & $Y$ is mid point of $AC.$

$\therefore O$ is mid point of $AK$(by mid point thm)

$\therefore AO=\frac{1}{2}AK$