Question

ABCD is a trapezium in which AB II CD. The diagonals AC and BD intersect at O. Prove that : (i) ΔAOB ~ ΔCOD

(ii) If OA = 6 cm, OC = 8 cm, Find:

(a) \(\frac{Area\,(ΔAOB)}{Area\,(ΔCOD)}\)

(b)  \(\frac{Area\,(ΔAOD)}{Area\,(ΔCOD)}\)

Answer 1

We have,

AB||DC 

In ΔAOB and ΔCOD ∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angle)

Then , ΔAOB ~ ΔCOD (By AA similarity) 

(a) By area of similar triangle theorem.

(b) Draw DP ⏊ AC