Question

In a given pentagon ABCDE, triangles ABC, BCD, CDE, DEA and EAB all have the same area. The
lines AC and AD intersect BE at points M and N. Prove that BM = EN.

Answer 1

Answer on Question #36915 – Math - Geometry

In a given pentagon ABCDE, triangles ABC, BCD, CDE, DEA and EAB all have the same area. The lines AC and AD intersect BE at points M and N. Prove that BM = EN.

Solution

$\triangle$ BCD and $\triangle$ CDE are of the same area. BCDE is trapezoid, where CD||BE

likewise, $\triangle BCD = \triangle ABC$ , so ABCD is trapezoid, where $BC \parallel AD$

for $\triangle CDE$ and $\triangle DEA$ DE|CA

from this we can say that BNDC and MCDE are parallelograms

now, $BN = CD = ME$

so, $BM = EN$