For an arbitrary vertex $A$ of an arbitrary triangle, using a compass, how can one find a point $p$ such that the line that goes through $A$ and $p$ divides the triangle into two pieces with the equal area? See the image for clarification.

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  • 4
    $\begingroup$ I dunno, I suppose one could try bisecting the side $BC$.... $\endgroup$ – Angina Seng Sep 5 '17 at 17:33
  • $\begingroup$ Bisecting the base is a solid idea. The two sides will have the same perpendicular height, but half the base length. $\endgroup$ – Theo Bendit Sep 5 '17 at 17:37
  • $\begingroup$ Using a compass but no straight edge? $\endgroup$ – Joffan Sep 5 '17 at 17:47

Bisect the side opposite $A$ and connect that point with $A$. Two triangles will have the same base and height, thus they will have same area. Do you know how to bisect a segment using a compass?

  • $\begingroup$ Bisection is perfect. Done in High school right. $\endgroup$ – Srijit Sep 5 '17 at 17:43
  • $\begingroup$ Bisecting a segment with only a compass is a challenge and it shouldn't be confused with using a compass and straightedge. To bisect a segment with only a compass look at this address: math.stackexchange.com/questions/227285/… $\endgroup$ – Seyed Sep 5 '17 at 18:37
  • $\begingroup$ @Seyed: it's definitely much easier to do the bisection with a straightedge in addition to compass $\endgroup$ – Vasya Sep 5 '17 at 19:38
  • $\begingroup$ @Vasya, It is true, but I only wrote that because straightedge was not mentioned in the question. $\endgroup$ – Seyed Sep 6 '17 at 0:12

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