Question

Prove that diagonal of a trape zium divide each other proportionally.

Answer 1

$<b> <i>$
Hey there !!

→ Given :- A trapezium ABCD in which AB || CD and its diagonals AC and BD intersect at point O.

$\bf { → To \: Prove :- \: \frac{AO}{OC} = \frac{BO}{OD}. }$

→ Construction :- Through O, draw EO || AB, meeting AD at E.

→ Proof :-

▶ In ∆ABC, EO || DC. [ => EO || AB || DC ].

$\bf { => \frac{AE}{ED} = \frac{AO}{OC} .......(1). [ by \: Thales' \: theorem ]. }$

▶ In ∆DAB , EO || AB.

$\bf { => \frac{DE}{EA} = \frac{DO}{OB} .....[ By \: Thales' \: theorem ]. }$

$\bf { => \frac{AE}{ED} = \frac{BO}{OD} . .....(2) }$

➡ From equation (1) and (2), we get

$\bf { => \frac{AO}{OC} = \frac{BO}{OD}. }$

✔✔Hence, it is proved ✅✅.

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$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

here is your answer OK


friend we knuw that one opposites sides are parallel AD||DC and diagonals also bisect each other

Given-A trap. ABCD in which AB||DC and its diagonals AC and BD intersect at o

TO prove-AO/OC=BO/OD

const. through o draw EO||ab meeting AD at E

Proof-in triangle adc EO||AB

SO,BY THALES THEOREM

AE/ED=AO/OC ......(1)

IN TRIANGLE DAB EO||AB

SO BY THALES THEOREM AE/ED=BO/DO......(2)

(SINCE DE/EA=DO/OB BY THALES THEOREM)

FROM 1 2 WE GET

AO/OC=BO/OD

THIS IMPLIES DIAGONALS OF TRAP. DIVIDE EACH OTHER PROPORTIONALLY

hence PROOF.