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 : $areaof△ADM=areaof△ACN$.

Answer 1

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-
$A=\frac{1}{2}\ast AC\ast NP$
Area of triangle ACN-
$A=\frac{1}{2}\ast AC\ast MD$
Since $MN\parallel AC$
MD=NP
AREA(ACN)=AREA(ACM) EQ 1
NOW, AREA(ACM)=
$A=\frac{1}{2}\ast AM\ast CF$
NOW, AREA(ADM)=
$A=\frac{1}{2}\ast AM\ast dE$
Since $AB\parallel CD$
CF=DE
area(ACM)=area(ADM) EQ2
FROM EQ 1 AND EQ2
area(ADM)=area(ACN)