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# the diagonals of a quadrilateral abcd intersect each other at the point o such that

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## The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO = CODO . Show that ABCD is a trapezium.

Click here👆to get an answer to your question ✍️ The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO = CODO . Show that ABCD is a trapezium. The diagonals of a quadrilateral ABCD intersect each other at the point O such that

Question BO AO ​ = DO CO ​

. Show that ABCD is a trapezium.

Medium Open in App

Updated on : 2022-09-05

Given:

Solution Verified by Toppr

The diagonals of a quadrilateral ABCD intersect each other at the point O such that

BO AO ​ = DO CO ​ i.e., CO AO ​ = DO BO ​

To Prove: ABCD is a trapezium

Construction:

Draw OE∥DC such that E lies on BC.

Proof: In △BDC,

By Basic Proportionality Theorem,

OD BO ​ = EC BE ​ ............(1) But, CO AO ​ = DO BO ​

(Given) .........(2)

∴ From (1) and (2) CO AO ​ = EC BE ​

Hence, By Converse of Basic Proportionality Theorem,

OE∥AB Now Since, AB∥OE∥DC ∴ AB∥DC

Hence, ABCD is a trapezium. Video Explanation Solve any question of Triangles with:-

Patterns of problems

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स्रोत : www.toppr.com

## The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO=CO/DO . Show that ABCD is a trapezium.

The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO=CO/DO . Show that ABCD is a trapezium. Basic Proportionality Theorem

The diagonals... Question

ABCD

intersect each other at the point

O such that AOBO=CODO . Show that ABCD is a trapezium. Open in App Solution

Step 1. Explaining the diagram. Let ABCD

AC and BD

intersects each other at

O such that, AOBO=CODO

Step 2. Showing

ABCD

is trapeziumConstruction-From the point

O , draw a line EO touching AD at E in such a way that, EO∥DC∥AB In ΔDAB, EO || AB

By using Basic Proportionality Theorem

DEEA=DOOB........................(i)

Also, given, AOBO=CODO

⇒AOCO=BODO [applying alternendo]

⇒COAO=DOOB [ [applying invertendo]

⇒DOOB =COAO..........................(ii)

From equation (i) and (ii), We have DEEA=COAO

Therefore, By applying converse of Basic Proportionality Theorem,

EO || DC Also

EO || AB ⇒ AB || DC.

ABCD

is a trapezium with

AB || CD.

Suggest Corrections 16 SIMILAR QUESTIONS

Q.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that

Q. The diagonals of a quadrilateral

A B C D

intersects each other at the point

O such that A O B O = C O D O . Show that A B C D is a trapezium.

Q. Question 10

The diagonals of a quadrilateral ABCD intersect each other at the point O such that

A O B O = C O D O

. Show that ABCD is a trapezium.

Q. The diagonals of a quadrilateral

A B C D

intersect each other at the point

O , such that A O B O = C O D O . Then A B C D is a trapezium.

Q.

A B C D

is a trapezium in which

A B | | D C

and its diagonals intersect each other at point

′ O ′ . Show that A O B O = C O D O . View More RELATED VIDEOS Basic Proportionality Theorem

MATHEMATICS Watch in App EXPLORE MORE

Basic Proportionality Theorem

Standard X Mathematics

स्रोत : byjus.com

## Ex 6.2, 10

Ex 6.2, 10 The diagonals of a quadrilateral ABCD intersect each other at the point O such that 𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂 . Show that ABCD is a trapezium Given: ABCD is a quadrilateral where diagonals AC & BD intersect at O & 𝐴𝑂/𝐵𝑂=𝐶𝑂/𝐷𝑂 To prove: ABCD is a trapezium Construction: Let us draw a lin Check sibling questions

## Ex 6.2, 10 - Chapter 6 Class 10 Triangles (Term 1)

Last updated at March 16, 2023 by Teachoo    This video is only available for Teachoo black users

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### Transcript

Ex 6.2, 10 The diagonals of a quadrilateral ABCD intersect each other at the point O such that 𝐴𝑂/𝐵𝑂 = 𝐶𝑂/𝐷𝑂 . Show that ABCD is a trapezium Given: ABCD is a quadrilateral where diagonals AC & BD intersect at O & 𝐴𝑂/𝐵𝑂=𝐶𝑂/𝐷𝑂 To prove: ABCD is a trapezium Construction: Let us draw a line EF II AB passing through point O. Proof: Given 𝐴𝑂/𝐵𝑂=𝐶𝑂/𝐷𝑂 ⇒ 𝐴𝑂/𝐶𝑂=𝐵𝑂/𝐷𝑂 Now, in ∆ 𝐴𝐷𝐵 EO II AB 𝐴𝐸/𝐷𝐸=𝐵𝑂/𝐷𝑂 ⇒ 𝐴𝐸/𝐷𝐸=𝐴𝑂/𝐶𝑂 Thus in Δ ADC, Line EO divides the triangle in the same ratio ∴ EO II DC Now, EO II DC But, we know that EO II AB ⇒ EO II AB II DC ⇒ AB II DC Hence, one pair of opposite sides of quadrilateral ABCD are parallel Therefore ABCD is a trapezium . Hence proved

Next: Ex 6.3 → 