ABCD is a trapezium in which . The diagonals AC and BD intersect at O. Prove that : (i)
(ii) If OA = 6 cm, OC = 8 cm, Find:
(a) (b)
We have,
AB||DC
In ΔAOB and ΔCOD (Vertically opposite angles)
<OAB = <OCD (Alternate interior angle)
Then , ΔAOB ~ ΔCOD (By AA similarity)
(a) By area of similar triangle theorem.
Area ΔAOB/area ΔCOD = OA2/OC2
Or, 62/82
Or, 36/64
Or, 9/16
b) Draw DP ⏊ AC
Area ΔAOB/area ΔCOD = 1/2 OA x DP/ 1/2 OC x DP
Or, OA/OC
Or, 6/8
Or, 3/4