Question

$\huge \red{Question - }$
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD.

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Answer 1

$\rm \leadsto \bigstar{Concept}$

Here, the above query said that A Trapezium ABCD having it's sides AB & CD are parallel to each other. And it's two diagonals AC & BD are interesting each other at O.

How to do?

Here, we have to prove that Area of△ AOB and Area of △COD is equal. There are total three triangle Similarity theorem . They are :- i) Angle - Angle - Angle ( AAA ) , ii) Side - Side - Side ( SSS ) , iii) Side - Angle - Side ( SAS ) . Here AAA triangle Similarity theorem is used.

$\rm \leadsto \bigstar{The \: Question \: is \: Given \: that}$

$\rm Diagonals \: AC \: and \: BD \: of a \: trapezium \quad \quad \\ \rm ABCD \: with \: AB//DC \: intersect \: each \: other \\ \rm at \: the \: point \: O. \: Using\: a \: similarity \: criterion \\ \rm for \: two \: triangles \: . \: Show \: the \: \frac{AO }{OC} = \frac{OB}{OD} .$

$\rm \leadsto \bigstar{Full \: Solution \: With \: Explaination}$

$\rm { \red{ \underline {Given \: that}}}$

$\rm ✦ ABCD \: is \: a \: trapezium \quad \quad \quad \quad \quad \quad \quad \: \: \: \: \: \: \: \: \: \: \\ \rm✦AB// AD \: \quad \quad \quad \quad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad \quad\quad \\ \rm✦ Diagonals \: AB \: and \: CD \: are \: intersecting \: at \: O$

$\rm { \red{ \underline {To \: Prove}}}$

$\rm \frac{AO}{OC} = \frac{OB}{OD}$

$\rm { \red{ \underline {Proof}}}$

$\rm In \: △\: AOB \: and \:△ \: COD$

$\rm i) \: ∠AOB = ∠ COD \: ( Vertically \: opposite \: angles \: are \: equal)$

$\rm ii) \: ∠OBA= ∠ ODC \: ( Alternate \: interior \: angles \: are \: equal)$

$\rm iii) \: ∠OAB = ∠OCD \: ( Alternate \: interior\: angles \: are \: equal)$

$\sf△AOB ∼△COD \: (By \: AAA \: Similarity)$

$\rm \leadsto \bigstar{Diagram}$

$\rm \: in \: the \: above \: attachment$

Answer 2

$\huge \red{ANSWER - }$

ABCD is a trapezium with AB∥CD and diagonals AB and CD intersecting at O.

⇒ In △OAB and △OCD

⇒ ∠AOB=∠DOC [ Vertically opposite angles ]

⇒ ∠ABO=∠CDO [ Alternate angles ]

⇒ ∠BAO=∠OCD [ Alternate angles ]

∴ △OAB∼△OCD [ AAA similarity ]

We know that if triangles are similar, their corresponding sides are in proportion.

$\frac{oa}{oc} = \frac{ob}{od}$ [ Hence, Proved]

❤Adyasha here❤