D ADRBC परस्परमा विन्दुमा
काटिएका छन् । प्रमाणित गर:
0 AAOBADOC
AO=DOBOCO
in the adjoining figure, AB||CD,
ABCD and AD and BC Intersect
each other to prove that
AAOB ADOC
(1) AO = DO and BO = CO​

what that's sorry I can't understand your questions

I can't understand

can't you English that's your questions?

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation:

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)