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Byju's Answer

Standard X

Mathematics

Pythagoras Theorem

In the follow...

Question

In the following figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC × BD

(ii) AC² = BC × DC

(iii) AD² = BD × CD

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Solution

(i) In ,

∴ (AA similarity criterion)

(ii)

(iii)

∠DCA = ∠ DAB (Each 90º)

∠CDA = ∠ ADB (Common angle)

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Similar questions

$Δ A B D$ is a right triangle right-angled at A and AC \perp BD. Show that
(i) $A B^{2} = B C . B D$
(ii) $A C^{2} = B C . D C$
(iii) $A D^{2} = B D . C D$
(iv) $\frac{A B^{2}}{A C^{2}} = \frac{B D}{D C}$

Q. In Fig.,

A B D

is a triangle right angled at

A

and

A C

⊥

BD. Show that:
(i)

A B^{2} = B C . B D

(ii)

A C^{2} = B C . D C

(iii)

A D^{2} = B D . C D

Q. Question 3 (i)
In the given figure, ABD is a triangle right angled at A and

A C ⊥ B D

. Show that

(i) A B^{2} = B C \times B D

△ A B D

is a right triangle right-angled at A and

A C ⊥ B D .

Show that

(i)

{A B}^{2} = B C . B D

(ii)

{A C}^{2} = B C . D C

(iii)

{A D}^{2} = B D \cdot C D

(iv)

\frac{{A B}^{2}}{{A C}^{2}} = \frac{B D}{D C}

Q. In the figure, ABC is triangle right angled at A and

A C ⊥ B D

Show that
(i)

A B^{2} = B C . B D

(ii)

A C^{2} = B C . D C

(III)

A D^{2} = B D . C D

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In the following figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB2 = BC × BD (ii) AC2 = BC × DC (iii) AD2 = BD × CD

(i) In , ∴ (AA similarity criterion) (ii) (iii) ∠DCA = ∠ DAB (Each 90º) ∠CDA = ∠ ADB (Common angle)

In the following figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC × BD

(ii) AC² = BC × DC

(iii) AD² = BD × CD

(i) In ,

∴ (AA similarity criterion)

(ii)

(iii)

∠DCA = ∠ DAB (Each 90º)

∠CDA = ∠ ADB (Common angle)