In figure- if - A - C -90- then prove that AOB - COD

In triangles AOB and COD, we have ∠ A = ∠ C ex[Given] and, ∠ 1 = ∠ 2 ex [Vertically opposite angles] Therefore, by AA criterion of similarity, we have ex AOB ∼ ...

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Standard X

Mathematics

AAA Similarity

In figure, if...

Question

In figure, if $∠ A = ∠ C = 90^{\circ}$ , then prove that $△ A O B \sim △ C O D$

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Solution

In triangles AOB and COD, we have
$∠ A = ∠ C$ [Given]
and, $∠ 1 = ∠ 2$ [Vertically opposite angles]
Therefore, by AA-criterion of similarity, we have
$△ A O B \sim △ C O D$

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Similar questions

Q. In given figure, If

∠ A = ∠ C

, then prove that

△ A O B \sim △ C O D

Q. In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 95°, then ∠COD = ?
(a) 95°
(b) 85°
(c) 90°
(d) 55°

Q. In the given figure, prove that

∠ A O B + ∠ B O C + ∠ C O D + ∠ D O A = 360^{\circ} .

In the following figure, OA =OC and AB =BC.

Prove that :

(i) $∠$ AOB = $90^{0}$

(ii) $Δ$ AOD $≅ Δ$ COD

(iii) AD =CD

Q. In the following figure,

O A = O C

and

A B = B C

. Prove that :
i)

∠ A O B = 90^{o}

ii)

Δ A O D ≅ Δ C O D

iii)

A D = C D

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In figure, if ∠A=∠C=90∘, then prove that △AOB∼△COD

In triangles AOB and COD, we have ∠A=∠C[Given] and, ∠1=∠2 [Vertically opposite angles] Therefore, by AA-criterion of similarity, we have △AOB∼△COD

In figure, if $∠ A = ∠ C = 90^{\circ}$ , then prove that $△ A O B \sim △ C O D$

In triangles AOB and COD, we have
$∠ A = ∠ C$ [Given]
and, $∠ 1 = ∠ 2$ [Vertically opposite angles]
Therefore, by AA-criterion of similarity, we have
$△ A O B \sim △ C O D$