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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4I dunno, I suppose one could try bisecting the side $BC$.... – Angina Seng Sep 05 '17 at 17:33
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Bisecting the base is a solid idea. The two sides will have the same perpendicular height, but half the base length. – Theo Bendit Sep 05 '17 at 17:37
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Using a compass but no straight edge? – Joffan Sep 05 '17 at 17:47
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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?
Vasili
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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: https://math.stackexchange.com/questions/227285/constructing-the-midpoint-of-a-segment-by-compass – Seyed Sep 05 '17 at 18:37
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@Seyed: it's definitely much easier to do the bisection with a straightedge in addition to compass – Vasili Sep 05 '17 at 19:38
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@Vasya, It is true, but I only wrote that because straightedge was not mentioned in the question. – Seyed Sep 06 '17 at 00:12