Given: ABCD is a quadrilateral in which AC and BD are diagonals and they intersect each other at O such that OB = OD.
To prove:
(i) ar (∆DOC) = ar (∆AOB)
(ii) ar (∆DCB) = ar (∆ACB)
(iii) DA || CB or ABCD is a parallelogram.
Construction: Draw DN ⊥ AC and BM ⊥ AC
Proof: (i) O is the mid-point of diagonal BD.
Therefore AO and CO will be the medians of ∆ABD and ∆BCD respectively.
ar (∆AOD) = ar (∆AOB) …(i)
and ar (∆COD) = ar (∆BOC) …(ii)
Reason: Median of a triangle divides it into two triangles equal in areas.
ar (∆AOD) + ar (∆COD) = ar (∆AOB) + ar (∆BOC)
⇒ ar (∆ACD) = ar (∆ABC)
⇒ \(\frac { 1 }{ 2 }\) AC x DN = \(\frac { 1 }{ 2 }\) AC x BM
⇒ DN = BM
Now in ∆CND and ∆ABM
AB = CD (given)
DN = BM
and ∠DNC = ∠BMA = 90°
⇒ ∆CND = ∆ABM (by R.H.S. congruence rule)
⇒ ar (∆CND) = ar (∆ABM) …(iii)
Similarly, in ∆DON and ∆BOM
DN = BM
DO = OB (given)
and ∠DNO = ∠BOM = 90° (each)
∆DON = ∆BOM (by R.H.S. congruence rule)
⇒ ar (∆DON) = ar (∆BOM) …(iv)
On adding (iii) and (iv), we get
ar (∆CND) + ar (∆DON) = ar (∆ABM) + ar (∆BOM)
⇒ ar (∆DOC) = ar (∆AOB)
(ii) ar (∆DOC) = ar (∆AOB)
On Adding ar (∆BOC) to both side, we get
ar (∆DOC) + ar (∆BOC) = ar (∆AOB) + ar (∆BOC)
⇒ ar (∆DCB) = ar (∆ACB)
(iii) Since A’s DCB and ACB are on the same base BC and equal areas, therefore they must lie between the same parallels
i.e. DA || CB …(v)
∆CDN = ∆ABM and ∆DNO = ∆BMO [by using relations (iii) and (iv)]
∠CDN = ∠ABM and ∠NDO = ∠MBO
On adding, we get
∠CDN + ∠NDO = ∠ABM + ∠MBO
⇒ ∠CDO = ∠ABO
But these are alternate angles.
Therefore CD || BA and DA || CB.
Hence, ABCD is a parallelogram.
