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Standard X

Mathematics

AAA Similarity

Diagonals AC ...

Question

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

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Solution

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]

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Similar questions

Q. 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}

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. 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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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

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]

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]