# How to find a point that divides a triangle into to pieces with equal area

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.

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

• Bisection is perfect. Done in High school right. – Srijit Sep 5 '17 at 17:43
• 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/… – Seyed Sep 5 '17 at 18:37
• @Seyed: it's definitely much easier to do the bisection with a straightedge in addition to compass – Vasya Sep 5 '17 at 19:38
• @Vasya, It is true, but I only wrote that because straightedge was not mentioned in the question. – Seyed Sep 6 '17 at 0:12