Question

ABCD is a trapezium in which AB||DC and it diagonals intersect each other at point O show that AO/BO=CO/DO

Answer 1

let us first prove AOB=DOB.
BAO=DCO (alternate opp. angle)
AOB=COD (O is common)

AOB=COD (AA axiom)

we get, AO/CO=BO/DO =AB/CD.

taking,AO/CO=BO/DO

crossmultiplying,
AO/BO=CO/DO. hence proved.

any other related questions u can ask

Answer 2

Answer :

$\bold{\star{\red{Given : }}}$

• In trapezium ABCD , AB||CD
• Both diagonal intersect at "O" point

$\bold{\star{\red{To Prove :}}}$

$=> \boxed{\bold{\frac{AO}{BO} = \bold{\frac{CO}{DO}}}}$

$\bold{\star{\red{Proof :}}}$

Basic proportionality theorem

=> If a line is parallel to one side of a triangle and intersecting the other two sides of the triangle, then the other two sides of the triangle are divided in the same ratio.

→In ΔADC

$=> \boxed{\bold{\frac{AO}{CO} = \bold{\frac{AG}{AD}}}}$.............1] by Basic proportionality theorem

→In ΔBDA

$=> \boxed{\bold{\frac{BO}{OD} = \bold{\frac{AG}{AD}}}}$.....................2] By Basic proportionality theorem

From 1] and 2] Equation

$=> \boxed{\bold{\frac{AO}{CO} = \bold{\frac{AG}{AD} = \bold{\frac{BO}{OD} = \bold{\frac{AG}{AD}}}}}}$

∴

$=> \boxed{\bold{\frac{AO}{CO} = \bold{\frac{BO}{OD}}}}$

Now Cross multiplication :-

∴

$=> \boxed{\bold{\frac{AO}{BO} = \bold{\frac{CO}{DO}}}}$

Hence proved