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


12