Question

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

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

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

(iii) ${AD}^2=BD\cdot CD$

(iv) $\frac{{AB}^2}{{AC}^2}=\frac{BD}{DC}$

Answer 1

$\left(i\right)$ In $△ADB$ and $△CAB$

$\angle DAB=\angle ACB$ [ Each ${90}^o$ ]

$\angle ADB=\angle CBA$ [ Common angle ]

$\therefore$ $△ADB\sim △CAB$

$\therefore$ $\frac{AB}{CB}=\frac{BD}{AB}$

$\therefore$ $A{B}^2=BC\times BD$ [ Hence proved ]

$\left(ii\right)$ Let $\angle CAB=x$ --- ( 1 )

In $△CBA$

$\angle CBA={180}^o-{90}^o-x$

$\therefore$ $\angle CBA={90}^o-x$ --- ( 2 )

Similarly in $\angle CAD$,

$\angle CAD={90}^o-\angle CAD$

$\therefore$ $\angle CAD={90}^o-x$ ---- ( 3 )

$\angle CDA={180}^o-{90}^o-({90}^o-x)$

$\angle CDA=x$ --- ( 4 )

In $△CBA$ and $△CAD$

$\angle CBA=\angle CAD$ [ From ( 2 ) and ( 3 ) ]

$\angle CAB=\angle CDA$ [ From ( 1 ) and ( 4 ) ]

$\therefore$ $△CBA\sim △CAD$ [ By AA similarity ]

$\therefore$ $\frac{AC}{DC}=\frac{BC}{AC}$

$\therefore$ $A{C}^2=BC\times DC$ ---- [ Hence proved ]

$\left(iii\right)$ In $△DCA$ and $△DAB$

$\angle DCA=\angle DAB$ [ Each ${90}^o$]

$\angle CDA=\angle ADB$ [ Common angle ]

$\therefore$ $△DCA\sim △DAB$

$\therefore$ $\frac{DC}{DA}=\frac{DA}{DB}$

$\therefore$ $A{D}^2=BD\times CD$ ---- [ Hence proved ]

$\left(iv\right)$ $\frac{A{B}^2}{A{C}^2}=\frac{BC\times BD}{BC\times DC}$

$\therefore$ $\frac{A{B}^2}{A{C}^2}=\frac{BD}{DC}$