Question

Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at the point O using similarity criterion show that OA/OC=OB/OD

Answer 1

Answer:

Step-by-step explanation:

Given : In trapezium ABCD , AB || DC. OC is the point of intersection of AC and BD.

To prove = OA/OC = OB/OD

Now, in ΔAOB and ΔCOD

∠AOB  = ∠COD    

[Vertically opposite angles]

∠OAB   = ∠OCD    

[alternate interior angles]

Then, ΔAOB∼ΔCOD

Therefore, OA/OC = OB/OD    

[Since, triangles  are Similar ,hence, corresponding Sides will be proportional]

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Answer 2

Question:

Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at the point O using similarity criterion show that OA/OC=OB/OD

Given:

• AB || DC
• OA = OC
• OB = OD

To prove :

• $\bf \dfrac{OA}{OC}= \bf \dfrac{OB}{OD}$

Proof :

In $\triangle$ AOB and $\triangle$ DOC

$\angle$A = $\angle$C ㅤ(alternate interior angles)

$\angle$B = $\angle$D ㅤ(alternate interior angles)

By AA criterion $\triangle$ AOB ~ $\triangle$DOC

$\bf \dfrac{OA}{OC} = \bf \dfrac {OB}{OD}$

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