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# Geometry

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Point D is the midpoint of median AM of triangle ABC. Point E is the midpoint of AB, and point T is the intersection of BD and ME. Find the area of triangle BEC if [ABC] = 14.

Sep 1, 2024

#1
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Point E is on the midpoint of AB and [ABC] is 14. Draw line EC. SInce ∆BEC has the same base and the same height has ∆AEC. [BCE] + [ACE] = [ABC]. So [BCE]= 7

Sep 1, 2024
#2
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$$\triangle{AEM}$$ and $$\triangle{B EM}$$ share the same base and the same height, so $$[\triangle{AEM}]$$ = $$[\triangle{BEM}]$$ and therefore,  $$[\triangle{BEM}]$$ = $$\frac{1}{2}[\triangle{BAM}]$$$$[\triangle{BAM}]$$ and $$[\triangle{CAM}]$$ also share the same base and the same height, so $$[\triangle{BAM}]$$ = $$[\triangle{CAM}]$$ and therefore,  $$[\triangle{BAM}]$$ = $$\frac{1}{2}\triangle{ABC}$$. Draw line $$EC$$.  $$\triangle{BEM}$$ and $$\triangle{CEM}$$ also share the same base and the same height, so $$[\triangle{BEM}]$$ = $$[\triangle{CEM}]$$ and therefore,  $$[\triangle{BEM}]$$ = $$\frac{1}{2}[\triangle{BEC}]$$. In the end, you get $$[\triangle{BEC}]=2[\triangle{CEM}]=2\cdot\frac{1}{2}[\triangle{BAM}]=2\cdot\frac{1}{2}\cdot\frac{1}{2}[\triangle{ABC}]=\frac{1}{2}[\triangle{ABC}]=\frac{1}{2}\cdot14=7$$.

Sep 1, 2024