× base × height
=

1 / 2

× BD × AO
And, area of triangle ACD is given byar (ΔACD) =

1 / 2

× base × height
=

1 / 2

× DC × AO
=

1 / 2

× BD × AO [From (1)] = ar (ΔABD)
Thus, ar (ΔACD) = ar (ΔABD) because BD = DC; and they have common altitude (AO), that is they are between the same parallels.1e:T5f7,

From B and C, draw perpendiculars to AE.
In ΔBPD and ΔCQD,
∠BPD = ∠CQD (Each is right angle)
∠BDP = ∠CDQ (Vertically opposite angles)
BD = CD (AD is median of ∆ABC)
∴ ΔBPD ≅ ΔCQD (AAS congruence rule)
⇒ BP = CQ (By CPCT) …(1)
We know that two congruent triangles have equal areas.
Therefore,
ar (ΔBPD) = ar (ΔCQD) …(2)
Now, in ΔBPA and ΔCQE,
∠BPA = ∠CQE (Each is right angle)
AB = CE (Given)
BP = CQ [From (1)]
∴ ΔBPA ≅ ΔCQE (RHS congruence rule)
We know that two congruent triangles have equal areas.
So, we have
ar (ΔBPA) = ar (ΔCQE) …(3)
Adding (2) and (3), we get
ar (ΔBPD) + ar (ΔBPA) = ar (ΔCQD) + ar (ΔCQE)
⇒ ar (ΔBAD) = ar (ΔCED)
Adding area (ΔBDE) on both the sides, we get
ar (ΔBAD) + ar (ΔBDE) = ar (ΔCED) + ar (ΔBDE).
This implies
ar (ΔABE) = ar (ΔBEC)
We know that triangles on the equal bases and havi