Mathematics
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 △ ADM = area of △ ACN.
Answer
We know that,
Area of triangles on the same base and between same parallel lines are equal.
From figure,
△ ADM and △ AMC lie on same base AM and between same parallel lines MB and DC.
∴ Area of △ ADM = Area of △ AMC ………..(1)
△ AMC and △ ACN lie on same base AC and between same parallel lines MN and AC.
∴ Area of △ AMC = Area of △ ACN ………..(2)
From equation (1) and (2), we get :
⇒ Area of △ ADM = Area of △ ACN.
Hence, proved that Area of △ ADM = Area of △ ACN.
Related Questions
In the given figure, D is mid-point of side AB of △ ABC and BDEC is a parallelogram.
Prove that :
Area of △ ABC = Area of // gm BDEC.
In the following figure, AC // PS // QR and PQ // DB // SR.
Prove that :
Area of quadrilateral PQRS = 2 × Area of quad.ABCD.
In the given figure, AD // BE // CF. Prove that :
area (△ AEC) = area (△ DBF)
In the given figure, ABCD is a parallelogram. BC is produced to point X. Prove that :
area (△ ABX) = area (quad.ACXD)
