Question

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac { AO }{ BO } = \frac { CO }{ D{ O }^{ \bullet } }$ Show that ABCD is a trapezium. [Class 10]

Answer 1

Step-by-step explanation:

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Answer 2

Given :–

In recrangle ABCD,

• $\frac {AO}{BO} = \frac {CO}{DO}$

To Prove :–

• ABCD is a trapezium.

Proof :–

→ $\frac { AO }{ BO } = \frac { CO }{ DO }$ [Given]

Replace the CO and BO.

BO is in the place of CO and CO is in the place of BO.

→ $\frac { AO }{ CO } = \frac { BO }{ DO }$

By vertically opposite angles (V.O.A),

→ ∠AOB = ∠COD

By SAS similarity theorem,

→ ∆AOB∼∆COD

And by the corresponding parts of similar triangles (CPCT),

→ ∠OAB = ∠OCD

But they are in the alternate position.

So,

→ AB || DC.

Hence,

ABCD is a trapezium.

!!Hence Proved!!