In the given figure, ABD is a triangle right angled at A and $AC \perp BD$. Show that $AC^2 = BC.DC$
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Given:

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

To do: 

We have to show that $AC^2=BC.DC$.

Solution:

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. 

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Updated on: 10-Oct-2022

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