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The triangle area formed by the center line of the triangle is equal to 3/4 of the triangle area.

Give a delta △ABC. The center line is CD, BF and AE. (as shown on the right)? Solution: connect DE, double the length to P, and connect BP, FP and EF. DE = EP，∠BEP=∠DEC，BE=EC。 ? ∴△DEC≌△PEB(SAS).？ ∴CD=BP.？ S△DEC=S△PEB。 ? And ∵DE parallel is equal to 1/2AC, and DE=EP. ∴EP is parallel and equal to 1/2AC. That is, EP is parallel and equal to AF. ∴ parallelogram AEPF. A parallelogram whose opposite sides are parallel and equal is a parallelogram? ∴AE=FP.？ S△EFP=S△AEF。 ? In this way, the three midline CD, BF and EF of △ABC constitute △BFP. ∵BF is the center line, and the area of △ABC is equally divided. ∴S△BAF=S△BFC.？ And ∵EF is the center line of △BFC, which divides the area of △BFC equally. ∴S△BEF=S△EFC= 1/4？ S△ABC。 ? And ∵CD is the center line of △ABC, and the area of △ABC is equally divided. ∴S△ADC=S△BDC.？ And de divides the area of △BDC equally. ∴S△BDE=S△DEC= 1/4？ S△ABC。 ? ∴S△BEP=S△DEC= 1/4？ S△ABC。 ? ∵AE is the center line of △ABC, which divides the area of △ABC equally. ∴S△BAE=S△AEC.？ And the average score of ÷EF △AEC. ∴S△AEF=S△EFC.？ ∴S△AFE=S△EFP= 1/4？ S△ABC？ ∫S△BFP = S△BEF+S△BEP+S△EFP？ = 1/4? S△ABC+ 1/4？ S△ABC+ 1/4？ S△ABC？ =3/4? S△ABC