+1439 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.
+2 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
+4 \(\triangle{AEM}\) 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 \([\triangle{BEC}]=2[\triangle{CEM}]=2\cdot\frac{1}{2}[\triangle{BAM}]=2\cdot\frac{1}{2}\cdot\frac{1}{2}[\triangle{ABC}]=\frac{1}{2}[\triangle{ABC}]=\frac{1}{2}\cdot14=7\).
