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In a $\triangle ABC$ If $P$ be a point which is Inside the $\triangle ABC$ such that

Area of$\triangle APB = $Area of $\triangle BPC = $Area of $\triangle CPA$.

Then how can I prove that the point $P$ is the centroid of $\triangle ABC$?

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Hints:

  • In $\triangle ABC$, let $M$ be the middle of $BC$, prove that $|\triangle ABM| = |\triangle ACM|$ (consider the heights perpendicular to $AM$).
  • Let $E$ be an interior point of $\angle BAC$. Prove that $|\triangle ABE| = |\triangle ACE|$ if and only if the point of intersection of $|AE|$ and $|BC|$ is $M$ (observe, that $|AE|$ is common to both and thus does not change the ratio).

Good luck ;-)

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Coordinates are not really necessary. If $$ [APB]=[APC], $$ then $P$ lies on the median from the $A$-vertex in the triangle $ABC$, since, if you put $P_A = AP\cap BC$, you have: $$1=\frac{[APB]}{[APC]}=\frac{[AP_A B]}{[AP_A C]}=\frac{BP_A}{CP_A}.$$ Since the argument works for all the vertices, $P$ lies in the intersection of the medians, i.e. $P$ is the centroid of $ABC$.

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If we are allowed to use Coordinate Geometry, let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ and $P(h,k)$

Now, calculate the area of the $3$ triangles using this (Article#25) and then equate them to form two equations of $h,k$ and solve for $h,k.$

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  • $\begingroup$ Thanks lab bhattacharjee $\endgroup$
    – juantheron
    Feb 17 '13 at 7:43

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