In the given figure, ABD is a triangle right angled at A and $AC \perp BD$. Show that
(i) $AB^2 = BC.BD$
(ii) $AC^2 = BC.DC$
(iii) $AD^2= BD.CD$
"

Given:

$\triangle ABD$ is a right triangle right-angled at A and $AC \perp BD$. 

To do: 

We have to show that

(i) $AB^2 = BC.BD$

(ii) $AC^2 = BC.DC$

(iii) $AD^2= BD.CD$

Solution:

(i) In $\triangle ABD$ and $\triangle ABC$,

$\angle DAB=\angle ACB=90^o$

$\angle B=\angle B$   (Common angle)

Therefore,

$\triangle ADB \sim\ \triangle CAB$   (By AA similarity)

This implies,

$\frac{AB}{CB}=\frac{BD}{AB}$   (Corresponding parts of similar triangles are proportional)

$AB.AB=CB.BD$   (By cross multiplication)

$AB^2=BC.BD$

Hence proved. 

(ii) Let $\angle CAB=x$,

This implies,

$\angle CAD=90^o-x$

In $\triangle CAB$,

$\angle CAB+\angle BCA+\angle ABC=180^o$

$x+90^o+\angle ABC=180^o$

$\angle ABC=180^o-90^o-x=90^o-x$

In $\triangle CAD$,

$\angle CAD+\angle CDA+\angle ADC=180^o$

$90^-x+90^o+\angle ADC=180^o$

$\angle ADC=180^o-180^o+x=x$

Therefore,

In $\triangle CAB$ and $\triangle CAD$,

$\angle CAB=\angle ADC=x$

$\angle ABC=\angle CAD=90^o-x$ 

Therefore,

$\triangle CAB \sim\ \triangle CDA$   (By AA similarity)

This implies,

$\frac{AC}{DC}=\frac{BC}{AC}$   (Corresponding parts of similar triangles are proportional)

$AC.AC=CB.DC$   (By cross multiplication)

$AC^2=BC.DC$

Hence proved. 

(iii) In $\triangle DCA$ and $\triangle DAB$,

$\angle DCA=\angle DAB=90^o$

$\angle CDA=\angle BDA$    (Common angle) 

Therefore,

$\triangle DCA \sim\ \triangle DAB$   (By AA similarity)

This implies,

$\frac{DC}{DA}=\frac{DA}{DB}$   (Corresponding parts of similar triangles are proportional)

$DA.DA=DB.DC$   (By cross multiplication)

$AD^2=BD.CD$

Hence proved. 

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Updated on: 10-Oct-2022

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