Question

M is the midpoint of seg AB and seg CM is a median of triangle ABC

State the reason

Answer 1

In the attachments I have answered this problem. See the attachments for detailed solution.

Answer 2

Answer:

Step-by-step explanation:

Let H be the height of the given triangle ABC.

It is given that M is the mid point of the segment AB and segment CM is the median of triangle ABC, therefore

AM=BM=$\frac{AB}{2}$

Now, Area of triangle= $\frac{1}{2}{\times}base{\times}height$

therefore, $\frac{ar({\triangle AMC})}{ar({\triangle BMC})}$=$\frac{\frac{1}{2}{\times}AM{\times}h}{\frac{1}{2}{\times}BM{\times}h }$

=$\frac{AM}{BM}$

=$\frac{\frac{AB}{2}}{\frac{AB}{2}}$

=$1$