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Consider a point P in the interior of a triangle ABC. Draw three lines passing through the point and parallel to the sides of the triangle.

They divide the triangle into three parallelograms and three triangles. Let S1, S2, and S3 be the areas of these smaller triangles. Find the area of the triangle ABC.

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This is what I have completed so far but I am really stuck. enter image description here

Ellie
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    hint: what is $S_1,S_2,S_3$ in terms of barycenteric coordinates of $P$. – achille hui Oct 04 '21 at 06:38
  • I know that S1, S2, and S3 are similar to the triangle ABC – Ellie Oct 04 '21 at 15:27
  • hint2: Let $S$ be area of $\triangle ABC$. If $(u,v,w)$ is barycentric coordinate of $P$ wrt $\triangle ABC$, then $u,v,w$ are ratios of heights $\implies u^2,v^2,w^2$ are ratios of areas. eg. $v$ = height of $\triangle IPE$ at $P$ : height of $\triangle ABC$ at $B \implies S_3 = Sv^2$. Now $u+v+w = 1$, this means... – achille hui Oct 05 '21 at 04:57
  • Ok I understand that and I proved that u + v + w = 1. But how do I use that to find the area of ABC? – Ellie Oct 05 '21 at 14:59
  • @Ellie the key is $P$ is an interior point, so $u,v,w > 0$. this allow you to sum the three $\sqrt{S_k}$ to get $\sqrt{S}$. see answer for more detail. – achille hui Oct 05 '21 at 16:04

2 Answers2

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Let $S$ be area of $\triangle ABC$.

Let $(u,v,w)$ be barycentric coordinate of point $P$ with respect to $\triangle ABC$.

By definition, $u,v,w$ are three numbers such that $$u+v+w = 1\quad\text{ and }\quad\vec{P} = u\vec{A} + v\vec{B} + w\vec{C}$$ Since $P$ lies in the interior of $\triangle ABC$, $u, v, w > 0$. Furthermore, the barycentric coordinates are ratio of heights of triangles. As an example,

$$v = \text{ height of } \triangle PAC \text{ at } P : \text{ height of } \triangle ABC \text{ at } B $$ Since $\triangle PIE$ is similar to $\triangle ABC$ and has same height as $\triangle PAC$ at $P$, we find

$$S_3 = \verb/Area/(\triangle PIE) = \verb/Area/(\triangle ABC) v^2 = S v^2$$ By a similar argument, we have $S_2 = S u^2$ and $S_1 = Sw^2$.

Since $u, v, w > 0$, we obtain

$$\sqrt{S} = \sqrt{S}(u+v+w) = \sqrt{Su^2} + \sqrt{Sv^2} + \sqrt{Sw^2} = \sqrt{S_2} + \sqrt{S_3} + \sqrt{S_1}$$

achille hui
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Similar triangles are to one another in the squared ratio of (their) corresponding sides[1]

Therefore, $$\small [\triangle FIP]:[\triangle PHE]:[\triangle GPG]=S_1:S_2:S_3$$ $$\small\implies FP:PE:GJ=\sqrt{S_1}:\sqrt{S_2}:\sqrt{S_3}$$

From that $\small AC:GJ=(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3}):\sqrt{S_3}$

As $\small \triangle ABC\sim\triangle GPJ\implies$ $$\small \begin{align} \small \frac{[\triangle ABC]}{[\triangle GPJ]}&=\frac{(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2}{(\sqrt{S_3})^2}\\ \small \frac{[\triangle ABC]}{S_3}&=\frac{(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2}{{S_3}}\\ \boldsymbol {[\triangle ABC]}&\boldsymbol{=(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2}\end{align}$$

ACB
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