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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 AB
lets 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=NP
AREA(ACN)=AREA(ACM) EQ 1
NOW, AREA(ACM)=
$$A=\frac { 1 }{ 2 } *AM*CF$$
NOW, AREA(ADM)=
$$A=\frac { 1 }{ 2 } *AM*dE$$
Since $$AB\parallel CD$$
CF=DE
area(ACM)=area(ADM)  EQ2
FROM EQ 1 AND EQ2
area(ADM)=area(ACN)

Mathematics

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