describe about mid point theoram and also describe about converse of mid poin theoram
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according to this theoram the line segment which joins the mid pts of two sides of a triangle is parallel and 1/2 of base of that triangle
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midpoint theorem is If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct . the other two sides are divided in the same ratio.
solution:-we are giving the triangle ABC in which a a line parallel two side BC intersect other two side a b and ac at D and e respectively:
we need to prove that AD/DB= AE/EC
let us join BE and CD and then draw DM perpendicular AC and EN perpendicular AB
now area of triangle ADE=1/2 base × height
1/2AD× EN
so,. ar(ADE)=1/2AD×EN
similarly,. ar(BDE)= 1/2 DB×EN
ar(ADE)=1/2 AE× DM and ar(DEC)=1/2 EC ×DM
therefore, ar(ADE)/ar(BDE)=(1/2AD×EN)/1/2DB×EN
and ar(ADE)/ar( BEC)=(1/2AE× DM)/1/2EC×DM
note that triangle BDE and DEC are on the same base and between the same parallels BC and DE.
so, ar(BDE)=ar(DEC)
therefore from 1 ,2 and 3 we have
AD/DB=AE/EC
solution:-we are giving the triangle ABC in which a a line parallel two side BC intersect other two side a b and ac at D and e respectively:
we need to prove that AD/DB= AE/EC
let us join BE and CD and then draw DM perpendicular AC and EN perpendicular AB
now area of triangle ADE=1/2 base × height
1/2AD× EN
so,. ar(ADE)=1/2AD×EN
similarly,. ar(BDE)= 1/2 DB×EN
ar(ADE)=1/2 AE× DM and ar(DEC)=1/2 EC ×DM
therefore, ar(ADE)/ar(BDE)=(1/2AD×EN)/1/2DB×EN
and ar(ADE)/ar( BEC)=(1/2AE× DM)/1/2EC×DM
note that triangle BDE and DEC are on the same base and between the same parallels BC and DE.
so, ar(BDE)=ar(DEC)
therefore from 1 ,2 and 3 we have
AD/DB=AE/EC
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