Given trapezoid ABCD with diagonals intersecting in X, find three pairs of triangles with equal areas. How would one go about doing and proving that?
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Originally Posted by Kolbeck10 Given trapezoid ABCD with diagonals intersecting in X, find three pairs of triangles with equal areas. How would one go about doing and proving that? ABD and ABC have equal base (AB) and altitude and hence equal area. CDB and CDA for the same reason have equal area. AXD and BXC, as ABD and ABC have the same area and: area(ABD)=area(AXD)+area(AXB) area(ABC)=area(BXC)+area(AXB). RonL
Originally Posted by CaptainBlack ABD and ABC have equal base (AB) and altitude and hence equal area. CDB and CDA for the same reason have equal area. AXD and BXC, as ABD and ABC have the same area and: area(ABD)=area(AXD)+area(AXB) area(ABC)=area(BXC)+area(AXB). RonL Is this dependent on the fact that it's an isosceles trapezoid, or would it be that way for one where the two non-parallel sides aren't the same?
Originally Posted by Kolbeck10 Is this dependent on the fact that it's an isosceles trapezoid, or would it be that way for one where the two non-parallel sides aren't the same? No, I tried to ignore that aspect of the diagram. RonL
Originally Posted by CaptainBlack No, I tried to ignore that aspect of the diagram. RonL Alright, perfect, thanks for the help.
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