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help plz

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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.

Dec 1, 2023

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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=21​BE=53​. Therefore, the area of triangle BET is 21​⋅BE⋅ET=21​⋅103​⋅53​=75​ square units.

Dec 1, 2023