Question

In $△ABC,\angle A$ is obtuse, $PB\perp AC$ and $QC\perp AB.$

Prove that:
(I) $AB\times AQ=AC\times AP$
(ii) ${BC}^2=(AC\times CP+AB\times BQ)$

Answer 1

In $△BPA$ and $△AQC$

$\angle P=\angle Q={90}^o$

$\angle PAB=\angle QAC$ [Vertically opposite angles]

and $\angle A$ is the common angle

$\therefore$ $△BPA\sim △AQC$ [AAA similarity criterion]

$\Rightarrow$ $\frac{AB}{AC}=\frac{AP}{AQ}$

$\therefore$ $AB\times AQ=AC\times AP$ --- ( 1 ) [ Hence proved ]

Consider, right angled $△BCQ$

$\Rightarrow$ $B{C}^2=C{Q}^2+B{Q}^2$ [By Pythagoras theorem]

$\Rightarrow$ $B{C}^2=C{Q}^2+(AB+AQ{)}^2$ [Since BQ = AB + AQ]

$\Rightarrow$ $B{C}^2=[C{Q}^2+A{Q}^2]+A{B}^2+2AB\times AQ$ ----- (2)

In right $△ACQ$, $C{Q}^2+A{Q}^2=A{C}^2$ [By Pythagoras theorem]

Hence equation (2) becomes,

$\Rightarrow$ $B{C}^2=A{C}^2+A{B}^2+AB\times AQ+AB\times AQ$

$\Rightarrow$ $B{C}^2=A{C}^2+A{B}^2+AB\times AQ+AP\times AC$ [From (1)]

$\Rightarrow$ $B{C}^2=A{C}^2+AP\times AC+A{B}^2+AB\times AQ$

$\Rightarrow$ $B{C}^2=AC(AC+AP)+AB(AB+AQ)$

$\Rightarrow$ $B{C}^2=AC\times CP+AB\times BQ$ [From the figure]

Hence Proved.