Question

# $$ABCD$$ is a trapezium with $$AB\: // \:DC$$ . A line parallel to $$AC$$ intersects $$AB$$ at point $$M$$ and $$BC$$ at point $$N$$ . Prove that : $$area \: of \: \bigtriangleup ADM \: = \: area \: of \: \bigtriangleup \: ACN$$.

Solution

## As per question,$$AC\parallel MN$$$$AB\parallel DC$$Lets DRAW NP, MD perpendicular to Ac,And ED perpendicular ABlets join A to C, D to M, C to M And A to N Area of triangle ACN-$$A=\frac { 1 }{ 2 } *AC*NP$$Area of triangle ACN-$$A=\frac { 1 }{ 2 } *AC*MD$$Since $$MN\parallel AC$$MD=NPAREA(ACN)=AREA(ACM) EQ 1NOW, AREA(ACM)=$$A=\frac { 1 }{ 2 } *AM*CF$$NOW, AREA(ADM)=$$A=\frac { 1 }{ 2 } *AM*dE$$Since $$AB\parallel CD$$CF=DEarea(ACM)=area(ADM)  EQ2FROM EQ 1 AND EQ2area(ADM)=area(ACN)Mathematics

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