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$\triangle ABD$ is a right triangle right-angled at A and $AC \perp BD$. Show that
$AC^2=BC.DC$
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.