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.

blackpanther Dec 1, 2023

#1**0 **

Since M is the midpoint of AM, we have that BM=AM=2AM, so B is the midpoint of AM. Hence, median AM is also an altitude, so △ABC is right-angled at M. Since △ABC is isosceles, we have that ∠A=∠C, and ∠B=180∘−∠A−∠C=90∘.

Let BE=a and BM=b. Since D is the midpoint of AM, we have that MD=2AM=2AB=10. Since D is also the midpoint of BM, we have that BD=2BM=b. Since ∠BME=∠BMD=45∘, triangles BME and BMD are congruent right triangles. Therefore, BE=BM=b, so E is also the midpoint of BM.

Since E is the midpoint of AB and M is the midpoint of AM, we have that AE=ME=2AB=10. By the Pythagorean Theorem, we have that BE=AB2−AE2=202−102=103. Since △BET is a 30-60-90 triangle with BE=103, we have that ET=21BE=53. Therefore, the area of triangle BET is 21⋅BE⋅ET=21⋅103⋅53=75 square units.

bader Dec 1, 2023