Measure EF aodWhar do you observe? You will find that:Esamplesee FigSo

1.	Measure EF aodWhar do you observe? You will find that:Esamplesee FigSolutionsa EFBCRepear this activity with some more triangles.Se, you arive at the following theorem:Fiheorem 89: The line segment joining the mid-points of two sidesparallel to the third side.d BCYo cae prove this theorgem using the followingObserve Fig &.25 in which E and Fare mid-points8 and AC respectively and CD II BAowÎ AEFr Î CDFDF and BE=AE = DC(ASA Rule)(Why?)EF:BL
Answer» The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it. Take a triangle ABC,E and F are the mid-points of side AB and AC resp. Construction:-Through C,draw a line II BA to meet EF produced at D. Proof:-In Triangle AEF and CDF1.AF=CF(F is midpoint of AC)2.<AFE=<CFD (Vertically opp. angles)3.<EAF=<DCF [Alt. angles,BA II CD(by construction) and AC is a transversal]4.So,Triangle AEF = CDF(ASA)5.EF=FD AND AE = CD (c.p.c.t)6.AE=BE(E is midpoint of AB)7.BE=CD(from 5 and 6)8.EBCD is a IIgm [BA II CD (by construction) and BE = CD(from 7)]9.EF II BC AND ED=BC (Since EBCD is a IIgm)10.EF = 1/2 ED (Since EF = FD,from 5)11.EF = 1/2 BC (Since ED = BC,from 9)Hence,EF II BC AND EF = 1/2 BC which proves the mid-point theorem.

Measure EF aodWhar do you observe? You will find that:Esamplesee FigSolutionsa EFBCRepear this activity with some more triangles.Se, you arive at the following theorem:Fiheorem 89: The line segment joining the mid-points of two sidesparallel to the third side.d BCYo cae prove this theorgem using the followingObserve Fig &.25 in which E and Fare mid-points8 and AC respectively and CD II BAowÎ AEFr Î CDFDF and BE=AE = DC(ASA Rule)(Why?)EF:BL

Answer»

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

Take a triangle ABC,E and F are the mid-points of side AB and AC resp.

Construction:-Through C,draw a line II BA to meet EF produced at D.

Proof:-In Triangle AEF and CDF1.AF=CF(F is midpoint of AC)2.<AFE=<CFD (Vertically opp. angles)3.<EAF=<DCF [Alt. angles,BA II CD(by construction) and AC is a transversal]4.So,Triangle AEF = CDF(ASA)5.EF=FD AND AE = CD (c.p.c.t)6.AE=BE(E is midpoint of AB)7.BE=CD(from 5 and 6)8.EBCD is a IIgm [BA II CD (by construction) and BE = CD(from 7)]9.EF II BC AND ED=BC (Since EBCD is a IIgm)10.EF = 1/2 ED (Since EF = FD,from 5)11.EF = 1/2 BC (Since ED = BC,from 9)Hence,EF II BC AND EF = 1/2 BC which proves the mid-point theorem.

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