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 triangles, show that $\frac{A O}{O C} = \frac{O B}{O D}$ .

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Solution

In $Δ D O C$ and $Δ B O A$ ,
$∠ C D O = ∠ A B O$ [Alternate interior angles as AB || CD]
$∠ D C O = ∠ B A O$ [Alternate interior angles as AB || CD]
$∠ D O C = ∠ B O A$ [Vertically opposite angles]
$∴ Δ D O C \sim Δ B O A$ [AAA similarity criterion]
$∴ \frac{D O}{B O} = \frac{O C}{O A}$ [Corresponding sides are proportional]
$\Rightarrow \frac{O A}{O C} = \frac{O B}{O D}$

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

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 triangles, show that $\frac{A O}{O C} = \frac{O B}{O D}$ .

Q. Diagonals

A C

and

B D

of a trapezium

A B C D

with

A B | | D C

intersect each other at the point

O .

\frac{O A}{O C} = \frac{O B}{O D}

Q. Diagonals AC and BD of a trapezium ABCD with AB

∥

DC intersect each other at the point O. Using similarity criterion for two triangles, show that

\frac{O A}{O C}

\frac{O B}{O D}

Q. In trapezium

A B C D

, side

A B ∥ C D

and diagonals

A C

and

B D

intersect each other at

O

Prove that:

\frac{O A}{O C} = \frac{O D}{O B}

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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 triangles, show that AOOC=OBOD.

In ΔDOC and ΔBOA, ∠CDO=∠ABO [Alternate interior angles as AB || CD] ∠DCO=∠BAO [Alternate interior angles as AB || CD] ∠DOC=∠BOA [Vertically opposite angles] ∴ΔDOC∼ΔBOA [AAA similarity criterion] ∴DOBO=OCOA [Corresponding sides are proportional] ⇒OAOC=OBOD

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 triangles, show that $\frac{A O}{O C} = \frac{O B}{O D}$ .