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I'm trying to tackle the following question, but unfortunately anything I tried got to a dead end (area of triangle, circles, angle bisectors, etc...) and I'm totally clueless how to solve it...

Let $P$ be some point on the side of given $\triangle ABC$ (WLOG, let $P$ be on BC).

Construct a line which passes through $P$ and halves the area of the triangle.

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Please give some hints. Thank you!

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  • $\begingroup$ Hint: Let $M$ be the midpoint of $BC$; then the line $AM$ halves the area of the triangle. Now construct a line $PQ$ through $P$ so that the area of $APM$ equals the area of $APQ$. $\endgroup$
    – user856
    Jul 3, 2016 at 11:48

2 Answers 2

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HINT : Draw a line $DE$ parallel to $AP$ passing through the midpoint $D$ of the side $BC$.

enter image description here

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  • $\begingroup$ Great hint, but I still couldn't mange to show that $PE$ halves the area. Could you please elaborate? $\endgroup$
    – Galc127
    Jul 3, 2016 at 12:28
  • $\begingroup$ @Galc127: The key is that $\triangle{ADP}=\triangle{AEP}$ (this is because $AP$ is parallel to $DE$). Using this gives that $\triangle{ACP}+\triangle{AEP}=\triangle{ACP}+\triangle{ADP}=\triangle{ACD}=(1/2)\triangle{ABC}$. Hence, $PE$ halves the area. $\endgroup$
    – mathlove
    Jul 3, 2016 at 12:38
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WLOG let $CP\ge BP$. Then reflect $C$ through $P$ and you will get $P'$. Now connect $P'$ and $A$ and draw parallel line to $AP'$, which intersects $AC$ at $X$. Then: $\left[\triangle CPX\right] = \frac{\left[\triangle ABC\right]}{2}$

To see why it holds not that from Intercept Theorem we have that: $$\frac{CA}{CX} = \frac{CP'}{CB} = \frac{2\cdot CP}{CB}$$

Hence:

$$\frac{\left[\triangle ABC\right]}{\left[\triangle CPX\right]} = \frac{CB \cdot CA \sin(\angle ACB)}{CP \cdot CX \sin (\angle ACB)} = \frac{CB}{CP} \times \frac{CA}{CX} = \frac{CB}{CP} \times \frac{2\cdot CP}{CB} = 2$$

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