Question

ABCD is a trapezium in wch AB || DC and its diagonals intersect each other at O.
Prove that: (AO/BO) = (CO/DO)

Answer 1

hy
here is your answer

given :-ABCD is a trapezium

where the AB ll DC

diagonals intersect each other at O

to prove:-(AO/BO) = (CO/DO)

construction :- let us draw a line EF ll AB ll DC

proof:- now in ΔADC

EO ll CD (because EF ll DC)

so, AE/DE=AO/CO..........(1) ( If a line is drawn parallel to one side of a triangle
to intersect the other two sides in distinct The other two sides are divided in the same ratio)


similarly in ΔDBA

EO ll. AB (EF ll AB)

so, AE/DE=BO/DO .........(2)( If a line is drawn parallel to one side of a triangle
to intersect the other two sides in distinct The other two sides are divided in the same ratio)
from (1)&(2)
AO/CO=BO/DO

AO/BO=CO/DO

hence proved

Answer 2

HEY , HERE'S THE ANSWER⏬

➡️ GIVEN : ABCD is a trapezium in which AB || DC and it's diagonal intersect each other at O.

➡️ CONSTRUCTION : Draw a line LO || DC || AB .

➡️ TO PROVE : (AO/BO) = (CO/DO)

➡️ PROOF : Let us consider ∆ADC -
(AL / LD) = (AO / OC) ----(i)
{ By THALES THEOREM }

Now , Similarly In ∆ADB -
( AL / LD ) = ( BO / DO )------(ii)

From eq. i and ii we get -

(AO/CO) = (BO/DO)
=> (AO/BO) = (CO/DO)

HOPE IT HELPS YOU ✌️!!!