Question

3. Diagonals AC and BD of a trapezium ABCD
with AB || DC intersect each other at the
point O. Using a similarity criterion for two
OB
ОА
triangles, show that
OC
OD
​

Answer 1

Answer:

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

Answer:

In ΔDOC and ΔBOA,

AB || CD, thus alternate interior angles will be equal,

∴∠CDO = ∠ABO

Similarly,

∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;

∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

Hence, proved.