Question

ABCD is a trapezium in which AB||CD . The diagonals AC and BD intersect at O. Prove that :
(i) ΔAOB~ ΔCOD
(ii) If OA = 6 cm, OC = 8 cm, Find:
(a) Area ΔAOB / Area ΔCOD (b) Area ΔAOD / Area ΔCOD

Answer 1

Use Similar Triangles Properties

Step-by-step explanation:

Given,

In the Trapezium ABCD,  

AB║CD and Diagonals AC intersects BD at O

To Prove: Congruency of ΔAOB and ΔCOD  

In both the triangles, (Refer Figure attached)

∠COD and ∠AOB (Vertically Opposite Angles, As AB║CD)

∠OAB = ∠OCD (As AB║CD, Alternate Interior Angles)

Hence, By AAA, Angle – Angle – Angle Similarity

∠AOB = ∠COD

A.) Using The Similar Triangle Theorem

$\frac {area AOB}{area COD} = \frac {OA^2}{OC^2} = \frac {6^2}{8^2} = \frac {36}{64} = \frac {9}{16}$

B.) Construction – Draw DX perpendicular to AC.

So, Again,  

$\frac {area AOD}{area COD} = \frac {\frac {1}{2} \times AO \times DP}{\frac {1}{2} \times CO \times DP} = \frac {AO}{CO} = \frac {6}{8} = \frac {3}{4}$