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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: enter image description here

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?

Anju George
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  • Hint: Yes, $AM_{A} = PM_{B}$. Look at triangles $\triangle CPM_B$ and $\triangle CA'A$. They are similar. – Sameer Kailasa Mar 09 '19 at 23:23
  • @SameerKailasa- Haha yes should've seen that. How about proving that the area of $BPM_B$ is equal to $3/8$ the area of $ABCA'$? Is it easy to see from this diagram? – Anju George Mar 09 '19 at 23:27
  • You will find very inspiring the answers of https://math.stackexchange.com/q/396085 – Jean Marie Mar 09 '19 at 23:29
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    Besides, use for example geogebra to draw your figures instead of a photograph of an approximate figure drawn on a blackboard :) – Jean Marie Mar 09 '19 at 23:31
  • See as well http://jwilson.coe.uga.edu/emt725/Medians.Triangle/Area.Medians.Tri.html – Jean Marie Mar 10 '19 at 09:03
  • You might want to have a look at https://gogeometry.blogspot.com/2016/12/geometry-problem-1296-herons-formula.html – Dr. Mathva Mar 10 '19 at 10:49

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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. enter image description here

Definition: $[...]$ denotes the area of the polygon "..."

Observe that $CE$ is a median in $\Delta CDH$. Thus

$$1=\frac{DK}{KH}\frac{[DKE]}{[EKH]}=\frac{[DKC]}{[KHC]}=\frac{[DKC]-[DKE]}{[KHC]-[EKH]}=\frac{[DEC]}{[EHC]}\iff [DEC]=[EHC]$$ Similarly $$[DEC]=[DHE]=[EHC]$$

Futhermore $$[DEC]=\frac{[ABC]}{4}$$ Can you end it now?

Dr. Mathva
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