Question

ABCD is a trapezium inwhich AB parallel to DC.CD is produced to E such that CE=AB. Prove that area of triangle ABD = area of triangle BCE.​

Answer 1

Step-by-step explanation:

Given that

ABCD is a trapezium in which AB parallel to DC.CD is produced to E such that CE=AB.

To prove

Area of triangle ABD = area of triangle BCE.

Proof

As

CE =AB

SO THEY LIE ON SAME BASES

Area of triangle ABD = area of triangle BCE.

HENCE PROVED....

HOPE IT HELPS AND PLZ MARK IT AS BRAINLIEST

Answer 2

Triangles between Over Same Base & Parallels Possess Equal Area

Step-by-step explanation:

Produce BA to M such that DM perpendicular BM and draw BN perpendicular DC.

Now,       ar(△ABD) = $\frac {1}{2} (AB \times DM)$            ....(i)

              ar(△BCE) = $\frac {1}{2} (CE \times BN)$               .....(ii)

 Since, ΔABD and ΔBCE are between the same parallels. Therefore,

                       DM = BN                              .........(iii)

Also,                AB = CE    (Given)            ............(iv)

    From (iii) and (iv), we get

         $\frac {1}{2} (AB \times DM) = \frac {1}{2}$ (CE \times BN)  

  ⇒      ar (△ABD) = ar (△BCE) (Using (i) and (ii))