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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

In the figure...

Question

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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Solution

ABD is a triangle right angled at A and
$A C ⊥ B D$
Prove

(i) Since $A C ⊥ B D$ hence $△ A B C \sim△ D C A$

and each triangle is similar to the whole
$△ A B D$

. $∵△ A B C \sim△ D B A$

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

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

(ii) Since $△ A B C \sim△ D A C$

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

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

(iii) Since $△ A C D \sim△ B A D$ ,

Hence $\frac{A D}{C D} = \frac{B D}{A D}$

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

Hence Proved.

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Similar questions

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

△ 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}

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

$Δ A B D$ 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 the given figure,

A B D

is a triangle right angled at

A

and

A C

perpendicular to

B D

.Show that

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

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

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