ABCD is a parallelogram. EF has been drawn parallel to BD such that it intersects the side AB at E and the side AD at F.

Prove that,

(i) the triangles BEC and DFC are equal in area.

(ii) the triangles AEC and AFC are equal in area

In the parallogram triangles ABC and ACD are equal in area because they're congruent:

AB=DC and AD=BC (opposite sides have equal length and both triangles ar between the same parallel lines AB and DC so they have equal area. Area of ACD=CDF+ACF and ABC=BCE+ACE. So (i) and (ii) are related by this commonality.

(i)

DF/AD=FG/AH (reduction in height CDF and ACD),
BE/AB=EQ/AP (reduction in height BCE and ABC).
Area BCE/area ABC=EQ/AP=BE/AB (triangle ABP),

Since the heights of the triangles (of equal area initially) are reduced by the same amount, their areas are also so reduced. This makes them equal in area.

In the picture, the heights of the triangles are shown by the perpendiculars FG, AH, EQ, AP.

For part (ii), having established (i) and knowing that the area of ACD=CDF+ACF and ABC=BCE+ACE, the areas of ACF and ACE are also equal.

answered Aug 27, 2016 by Top Rated User (425,020 points)

I didn't get your answer. Ok ABC area = ADC area. Then DFC+AFC=BEC+AEC. But from that how can we say DFC area =BEC area and AFC area=AEC area?

Finally I found a way to solve this math. Thank you.

Sorry for the delay, math93. I'm pleased you found a solution. Hope it wasn't as complicated as mine!

Thank you so much again for your answer. This is how I solved it.

Triangles on the same base and between the same pair of parallel lines are equal in area. So,

BEF Δ Area = DEF Δ Area  (On the same base EF and EF//BD.)

And also,

BED Δ Area = BEC Δ Area  (On the same base BE and AB//CD.)             BFD Δ Area = DFC Δ Area  (On the same base FD and AD//BC.)

But,

BEFD Area - BEF Δ Area = BEFD Area - DEF Δ Area                                                         So BED Δ Area = BFD Δ Area

So Because of that Finally,                                                                                                                    BEC Δ Area = DFC Δ Area

For Second Part,

ABC Δ Area = ADC Δ Area (Because they're congruent)

ABC Δ Area - BEC Δ Area ADC Δ Area - DFC Δ Area

So  AEC Δ Area = AFC Δ Area.

Yes, I like it!