This is Question $2$ from this document on Olympiad Geometry.
Let $ABC$ be a triangle and $M_A,M_B,M_C$ the midpoints of the sides $BC, CA, AB$, respectively. Show that the triangle with side lengths $AM_A, BM_B, CM_C$ has area $3/4$ that of the triangle $ABC$.
This is part of a chapter that stresses that by "slicing and dicing", we can solve a lot of complicated problems. Hence, the stress is on diagrammatic proofs.
To form a triangle with the medians, I extended $AM_A$ beyond $BC$, and formed another copy of the triangle $ABC$. My diagram looks like this:
Obviously $BP=CM_C$. Hence, if $PM_B=AM_A$, we'll have created a triangle with the medians as sides.
So my question is, is $AM_A=PM_B$? A followup question would be is it easy to see that the area of the triangle $BPM_B$ is equal to $3/8$ that of the parallelogram given?
Best Answer
Regarding the question, whether $AM_A=PM_B$, the answer is YES.
Simply observe that $$\frac{CP}{CA'}=\frac{CM_B}{CA}$$ Thus, in virtue of the Intercept theorem (also known as "Thales' Theorem") $$\frac{PM_B}{AA'}=\frac{CP}{CA'}=\frac{1}{2}$$
Can you end it now?
Alternatively, here you have a proof (almost) without words.
Definition: $[...]$ denotes the area of the polygon "..."
Observe that $CE$ is a median in $\Delta CDH$. Thus
Futhermore $$[DEC]=\frac{[ABC]}{4}$$ Can you end it now?