Triangle ABC has area 60 cm^2.Let AD be the median from A on BC and BY be median from B on AD. If BY is extended to meet AC in E what is the area of triangle AYE?


Using Menelaus' theorem, we have $$\frac{CB}{BD}\cdot\frac{DY}{YA}\cdot \frac{AE}{EC}=1\iff CE:EA=2:1.$$

Hence, the answer is $$60\times \frac{CD}{CB}\times\frac{EA}{CA}\times \frac{AY}{AD}=60\times\frac 12\times\frac 13\times\frac 12=5\ (\text{cm}^2).$$

  • $\begingroup$ Sorry can you explain it to me without vector method... Please $\endgroup$ – user2131465 Aug 10 '14 at 15:27
  • $\begingroup$ @user2131465: Sure. Take a look. I hope this helps. $\endgroup$ – mathlove Aug 10 '14 at 15:30
  • $\begingroup$ No, I still don't get it. I am unaware of the above theorem, I read it but I still don't get the second part, where we have CD/CB. $\endgroup$ – user2131465 Aug 10 '14 at 15:38
  • $\begingroup$ @user2131465: OK, let me expalin a bit. $60=ABC$ (here $ABC$ means the area of the triangle $ABC$.) and $60\times (CD/CB)=ADC$. And $ADC\times (EA/CA)=AED$. And $AED\times (AY/AD)=AYE$, which is what we need. $\endgroup$ – mathlove Aug 10 '14 at 15:41
  • $\begingroup$ @user2131465: Isn't my comment helpful? $\endgroup$ – mathlove Aug 10 '14 at 17:21

Menelaus theorem made easy

Since any affine map preserves the ratio between the areas, you are free to assume that the triangle $ABC$ has a right angle in $B$, and $$ A=(0,0),\quad B=(0,1),\quad C=(1,1),$$ hence: $$ D=(1,1/2),\qquad Y=(1/2,1/4).$$ Now it is not difficult to check that the $BY$ line meet the $AC$ line in the point $(1/3,1/3)$, hence: $$[AYE]=\frac{1}{3}[AYC]=\frac{1}{3}\cdot\frac{1}{2}[ACD]=\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}[ABC]=\frac{1}{12}[ABC]=5.$$


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