Question

Please answer it fastly.​

Answer 1

Answer:

Given : DE II OB

EF II BC

To prove : DF II OC

Construction : Join DF

Step-by-step explanation:

In triangle AOB and tri. AOC,

AO = AO [ common]

∠OBA = ∠OCA

AO is angle bisector,

So ∠BAO = ∠CAO

ΔABO ≅ ΔACO [ ASA similarity]

OB = OC [ by cpct] _ eq 1.

Like that we can prove ΔEDX ≅ ΔFDX [X is the point ]

EX = DX [ by cpct ] _ eq.2

Then DF II OC .[ from eq. 1 and eq 2 ]

Hope it helps you dear..