In triangle $\triangle LMN$. $LO$ is median. Also $LO$ is the bisector of angle. If $LO=3\text{ cm}$ and $LM=5\text{ cm}$. Then find the area of triangle $\triangle LMN$.
Now here we will use similarity between $\Delta LMO$ and $\Delta LON$ triangles but I could not proceed further.
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$\begingroup$Since $LO$ is both median and the bisector of $\angle NLM$, $\triangle LMN$ is isosceles and $ LO \perp MN$.
So $MO = NO= \sqrt{5^2-3^2\ }=4$ and area $\triangle LMN = 4\times 3 =12 \text{cm}^2$
To see that $\triangle LMN$ is isosceles, consider that $\angle LON = 180°-\angle LOM$. Flip $\triangle LON$ to superimpose $\angle NLO$ and $\angle MLO$. Then the only way to make $|MO|=|NO|$ is to have $\angle LON = \angle LOM =\perp$
$\endgroup$ 2 $\begingroup$Its' isosceles and acute. Because a acute is smaller than a right angle, and there is two of the sides are the same.
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