# The problem of congruent areas in a triangle.

A problem was posed in front of me and I couldn't solve it after multiple attempts--

Consider any triangle and 3 concurent cevians are drawn from each of its 3 points . Now the figure formed has 6 sub triangles - if the areas of 3 alternate sub triangles are equal then prove that the point of concurrence is the Centriod.

• what is a "cevian"? – enzotib Jul 9 '16 at 15:41
• @enzotib en.wikipedia.org/wiki/Cevian – Joffan Jul 9 '16 at 15:43
• I'm trying to understand exactly which areas you expect to be equal. The areas of the three shaded triangles are equal, or the shaded area vs. the unshaded area? – Joffan Jul 9 '16 at 15:45
• I believe the areas of the 3 shaded region ls is equal. – Shivam Patel Jul 9 '16 at 15:56
• "I believe the areas of the 3 shaded region ls is equal". Is equal to what? – fleablood Jul 9 '16 at 16:03 Assume that $\frac{BP_A}{P_A C}=\lambda$ and $\frac{CP_B}{P_B A}=\mu$. By Ceva's theorem $\frac{AP_C}{P_CB}=\frac{1}{\mu\lambda}$. Moreover: $$[BPP_A] = \frac{BP_A}{BC}[BPC]=\frac{BP_A}{BC}\cdot \frac{CP_B}{CA}[ABC]$$ hence $$[BPP_A]=\frac{\lambda}{\lambda+1}\cdot \frac{\mu}{\mu+1}[ABC].$$ In a similar way you may compute $[CPP_B]$ and $[APP_C]$ in terms of $\lambda,\mu,[ABC]$ and check that from $[BPP_A]=[CPP_B]=[APP_C]$ it follows that $\mu=\lambda=1$, i.e. $P\equiv G$.

• D' Aurizio I did not understand how $[BPP_A] = \frac{BP_A}{BC}[BPC]$ .Can you please explain? – Love Everything Jul 9 '16 at 16:58
• @LoveEverything: the ratio between $[BPP_A]$ and $[CPP_A]$ is the same as the ratio between $BP_A$ and $CP_A$: we have two triangles with the same height. – Jack D'Aurizio Jul 9 '16 at 16:59

Let $P$ be the intersection point inside $\triangle{ABC}$.

Let $D$ be the intersection point of $AP$ with $BC$, and let $E,F$ be the point on $AP$ such that $AE\perp BE, AF\perp CF$ respectively.

Since $\triangle{APB}=\triangle{APC}$, we have $BE=CF$ from which we know that $\triangle{BED}$ and $\triangle{CFD}$ are congruent, and so $BD=CD$. Hence, we know that $AP$ is the median.

Also, from $\triangle{PBD}=\triangle{PCD}$, we have that $$\triangle{PBC}=2\triangle{PBD}.$$

From $\triangle{PAB}=\triangle{PBC}$, we have that $$\triangle{PAB}=2\triangle{PBD}$$ from which $AP:PD=2:1$ follows.

It follows from this that $P$ is the centroid.

• Is it possible for you to please add a labeled figure? I (a mere high schooler) am having a little bit of trouble constructing a labeled figure as per your description. – Daniel Jul 9 '16 at 21:02