Question

Please answer my Question

Answer 1

Answer:

Since BC and DE are parallel, the triangles ABC and ADE are similar, so ABC is just a scaled version of ADE.  Call this scale factor k.

The perpendicuar distance from E to AD is then scaled by k to get the perpendicular distance from C to AB.  Since triangles ADC and ADE have the same base AD, but the height of ADC is k times the height of ADE, we get

area(ADC) = k times area(ADE)

Now look at triangles ADE and ABE, this time with AE as the "base".   Because of the fact that ABC is just ADE scaled by a factor of k, the perpendicular distance from B to AE is k times the perpendicular distance from D to AE.  So just as before, we get

area(ABE) = k times area(ADE).

Putting the two equations together, we have

area(ABE) = area(ADC),

as required.

Answer 2

ar(ECB)=ar(DCB) (Because both triangles lie on same base BC and between same parallel BCIICD)
Adding ACB on both sides
ar(ECB + ACB)=ar(DCB+ACB)
are(ABE)=are(ACD)

HENCE PROVED