Question

Plzz answer fast (Q.4)

Answer 1

here your answer....

Answer 2

Answer:

This is a very simple question of similarity.

In the given trapezium ABCD, AB║CD .

So, in Δ OMC and ΔONA,

∠DCA = ∠NAO ( alternate interior angles formed between two parallel lines AB and CD with AC as transverse line)

∠MOC = ∠NOA ( vertically opposite angles)

Now, in a triangle, sum of all the three angles is 180°.

So, in ΔOMC , ∠MOC + ∠OCM + ∠CMO = 180°

or, ∠ CMO = 180° - ∠MOC - ∠OCM.........................equation 1

Similarly, in ΔOAN, ∠OAN + ∠AON + ∠ONA = 180°

or, ∠ONA = 180° - ∠OAN - ∠AON

or, ∠ ONA = 180° - ∠DCA - ∠MOC .......................from equations 1 and 2

Hence, ∠ONA = ∠CMO.

Hence, by AAA similarity criterion, ΔOMC is similar to ΔONA.