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If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

AX, BY and CZ...

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

\frac{1}{4}

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\frac{1}{2}

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\frac{1}{3}

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\frac{2}{3}

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Solution

The correct option is B $\frac{1}{2}$ AK
Since angles BXO and CXK are similar.
$∴ \frac{O X}{X K} = \frac{B X}{O X} = 1$
$\Rightarrow O X = X K$
Also, $\frac{A O}{X K} = \frac{A O}{O X} = \frac{2}{1}$
$\Rightarrow A O = O K$
Hence, $A K = 2 A O$
$\Rightarrow A O = \frac{1}{2} A K$

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