In Fig. 3, $\angle ACB = 90^{o}$ and $CD  \perp AB$, prove that $CD^{2}= BD ×AD$.

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Given: $\angle ACB = 90^{o}$ and $CD  \perp AB$

To do: Prove that $CD^{2}= BD \times AD$.

Given that $CD \perp AB$.

$\angle ACB=90^{o}$

Using Pythagoras Theorem in $\vartriangle ACD$

$( AC)^{2}=( AD)^{2}+( CD)^{2}$    ..............$( i)$

Using Pythagoras Theorem in $\vartriangle CDB$,

$( CB)^{2}=( BD)^{2}+( CD)^{2} $   ..............$( ii)$

Similarly in $\vartriangle ABC$,

$( AB)^{2}=( AC)^{2}+( BC)^{2}$   ................$( iii)$

As $AB =AD+BD$          ................ $(iv)$

Squaring both sides of equation $( iv)$,

We get 

$( AB)^{2} =( AD+BD)^{2}$

Since, $AB^{2}=AD^{2}+BD^{2}+2\times BD\times AD$

From equation $( iii)$ we get

$AC^{2}+BC^{2}=AD^{2}+BD^{2}+2\times BD\times AD$

Substituting the value of $AC$ from equation $( i)$ and the value of $BC$ from equation $( ii)$, we get

$( AD)^{2}+( CD)^{2} +( BD)^{2}+( CD)^{2}=AD^{2}+BD^{2}+2\times BD\times AD$

$\Rightarrow 2CD^{2}=2\times BD\times AD$

$\Rightarrow CD^{2}=BD\times AD$

Hence, proved that $CD^{2}=BD\times AD$.