+269 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.
+1066 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.
