Mathematics
In the following figure, CE is drawn parallel to diagonal DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that △ ADE and quadrilateral ABCD are equal in area.

Theorems on Area
9 Likes
Answer
We know that,
Triangles on the same base and between the same parallel lines are equal in area.
△ BDE and △ BDC lie on the same base BD and along the same parallel lines DB and CE.
∴ Area of △ BDE = Area of △ BDC ……….(1)
From figure,
⇒ Area of △ ADE = Area of △ ADB + Area of △ BDE
⇒ Area of △ ADE = Area of △ ADB + Area of △ BDC [From equation (1)]
⇒ Area of △ ADE = Area of quadrilateral ABCD.
Hence, proved that △ ADE and quadrilateral ABCD are equal in area.
Answered By
4 Likes
Related Questions
In parallelogram ABCD, P is a point on side AB and Q is a point on side BC. Prove that :
(i) △ CPD and △ AQD are equal in area.
(ii) Area (△ AQD) = Area (△ APD) + Area (△ CPB)
In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.
If the area of parallelogram ABCD is 48 cm2;
(i) state the area of the triangle BEC.
(ii) name the parallelogram which is equal in area to the triangle BEC.

ABCD is a parallelogram, a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.

The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.
