In Fig. 6.53, 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
(i) $AB^2$ = BC . BD
In ΔADB and ΔCAB
$\angle DAB = \angle ACB $(Each 90°)
$\angle ABD = \angle CBA $(Common angles)
ΔADB ~ ΔCAB [AA similarity criterion]
$\dfrac{AB}{CB} = \dfrac{BD}{AB}$
$AB^2$ = CB × BD
(ii) $AC^2$ = BC . DC
Let $\angle CAB$ = x
In ΔCBA
$\angle CBA$ = 180° – 90° – x
$\angle CBA$ = 90° – x
Similarly, in ΔCAD
$\angle CAD = 90° – \angle CBA$
= 90° – x
$\angle CDA$ = 180° – 90° – (90° – x)
$\angle CDA$ = x
In ΔCBA and ΔCAD, we have
$\angle CBA = \angle CAD$
$\angle CAB = \angle CDA$
$\angle ACB = \angle DCA$ (Each 90°)
ΔCBA ~ ΔCAD [AAA similarity criterion]
$\dfrac{AC}{DC} = \dfrac{BC}{AC}$
$AC^2$ = DC × BC
(iii) $AD^2$ = BD . CD
In ΔDCA and ΔDAB
$\angle DCA = \angle DAB$ (Each 90°)
$\angle CDA = \angle ADB$ (common angles)
ΔDCA ~ ΔDAB [AA similarity criterion]
$\dfrac{DC}{DA} = \dfrac{DA}{DA}$
$AD^2$ = BD × CD
