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.

enter image description here

Please give some hints. Thank you!

  • $\begingroup$ 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


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

enter image description here

  • $\begingroup$ 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

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: $$\frac{CA}{CX} = \frac{CP'}{CB} = \frac{2\cdot CP}{CB}$$


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.