Question

In an isosceles $△$ ABC, the base AB is produced both the ways to P and Q such that $AP\times BQ={AC}^2$. Prove that $△APC\sim △BCQ$.

Answer 1

$\Rightarrow$ In given figure, $ABC$ is an isosceles triangle having

$AB$ and $AC=BC$. Also, $AP\times BQ=A{C}^2$

$\Rightarrow$ $\frac{AP}{AC}=\frac{AC}{BQ}$

$\Rightarrow$ $\frac{AP}{AC}=\frac{BC}{BQ}$ ------ ( 1 )

$\Rightarrow$ $AC=BC$ [ Given ] ----- ( 2 )

$\Rightarrow$ $\angle CAB=\angle CBA$ [ Angles opposite to equal sides are equal ]

$\Rightarrow$ $\angle CAB+\angle CAP={180}^o$ and $\angle CBA+\angle CBQ={180}^o$ [ Both forming a linear pair ]

$\Rightarrow$ $\angle CAB={180}^o-\angle CAP$ and $\angle CBA={180}^o-\angle CBQ$

$\Rightarrow$ ${180}^o-\angle CAP={180}^o-\angle CBQ$ [ From ( 2 ) ]

$\therefore$ $\angle CAP=\angle CBQ$

$\Rightarrow$ Now, in $△APC$ and $△CBQ$,

we have

$\Rightarrow$ $\angle CAP=\angle CBQ$

$\Rightarrow$ and $\frac{AP}{AC}=\frac{BC}{BQ}$

$\therefore$ $△APC\sim △CBQ$ [ SAS similarity ]+