Question

Diagonals AC and BD of a trapezium ABCD WITH AB || CD intersect each other at the point O. Using a similarity criteration for two triangles, show that OA/OC=OB/OD.​

Answer 1

Answer:

Proved!

Explanation:

Given Problem:

Diagonals AC and BD of a trapezium ABCD WITH AB || CD intersect each other at the point O. Using a similarity criteration for two triangles, show that OA/OC=OB/OD.

Solution:

To Prove:

OA/OC=OB/OD.​

Given that,

ABCD is a trapezium with AB║CD and diagonals AB & CD intersecting at O

Proof:

In ΔOAB and ΔOCD

∠AOB = ∠DOC    ★Vertically Opposite angle★

∠ABO = ∠CDO

★Since,AB║CD with BD as traversal and alternate angles★

∠BAO = ∠OCD

★Since,AB║CD with AC as traversal and alternate angles★

ΔOAB ~ ΔOCD

We know that,

★If 2 triangles are similar their corresponding sides are in proportion.★

Hence,

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

Hence Proved!

Note:

Refer the Attachments for figures.

Answer 2

Questions—

Diagonals AC and BD of a trapezium ABCD WITH AB || CD intersect each other at the point O. Using a similarity criteration for two triangles, show that OA/OC=OB/OD.

$\bf\huge\boxed{Answers}$

Given—

ABCD is a trapezium with AB || CD & Diagonals AB & CD intersecting at 'O'.

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To Prove—

OA/OC = OB/OD

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Proof—

A/Q

In ΔAOB & ΔDOC

∠BAO = ∠OCD ______(Alternate Angles)

∠ABO = ∠ODC ______(Alternate Angles)

∠AOB = ∠DOC ______(Vertically opposite angles)

Therefore, By AAA criterion of similarity, 

ΔAOB ~ ΔDOC

Therefore, OA/OC = OB/OD

_____(By Corresponding Sides of Similar Triangles)

HENCE PROVED..

☺☺☺THNXX☺☺☺✨