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Point $D$ is the midpoint of median $\overline{AM}$ of triangle $ABC$. Point $E$ is the midpoint of $\overline{AB}$, and point $T$ is the intersection of $\overline{BD}$ and $\overline{ME}$.  Find the area of triangle $BET$ if $AB = 20$, $AC = 20$, and triangle $ABC$ is isosceles.

 May 4, 2024
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Since triangle ABC is isosceles with AB=AC=20, segment AM is a median and an altitude (since the base angles are congruent). Therefore, △ABM is a right triangle with hypotenuse AB=20 and leg BM=21​⋅AB=10

 

Because D is the midpoint of AM, MD=21​⋅AM=21​⋅AB=10. Then △BDM is a 3-4-5 right triangle, so BD=4⋅10=40.

Similarly, since E is the midpoint of AB, AE=21​⋅AB=10. Then △AEM is also a 3-4-5 right triangle, so ME=4⋅10=40.

Triangle BET shares altitude EM with △AEM, and the ratio of their bases is AEBE​=21​. Therefore, the area of △BET is half the area of △AEM, which is 21​⋅21​⋅AE⋅EM=21​⋅21​⋅10⋅40=100​.

 May 4, 2024

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