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See complete Problem 89Triangle area, Midpoints. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Let X be the mid-pt of BC,XM//BA and XN//CD,S_1=[EBX]=S_3/2=[EXC]=S2
let area of ABC be Sarea of CEN = EC/ED*[END][END]= 1/2[BED][BED]=BE/AE*SThus, [CEN]=1/2*S*(EC/ED)*BE/AE[BEM]=BE/AE*[AEM][AEM]=1/2*[AEC][AEC]=EC/ED*SThus,[BEM]=1/2*S*(EC/ED)*(BE/AE)=[CEN][BEC]=(BE/AE)*(EC/CD)*STherefore, S1=S2=1/2S3
Here is an alternative solution to problem 89.Let CP and MQ be altitudes of triangles BEC and EBM respectively. In triangle APC, M and Q are midpoints of the sides, so MQ = CP/2,hence S1 = S3/2.Let BR and NS be altitudes of triangles BEC and ECN respectively. In triangle DBR, N and S are midpoints of the sides, so NS = BR/2,hence S1 = S2/2.Thus S1 = S2 = S3/2.
Area triangle MBE is (MBE)=S1=(AEM)-(ABM)=(AEC)/2-(ABC)/2=(BEC)/2=S3/2.But S2=(NEC)=(NED)-(NCD)=(BED)/2-(BCD)/2=(BEC)/2=S3/2.