Question

In $\Delta$ABC, $\angle$A is obtuse, PB$\perp$AC and QC$\perp$AB. Prove that $B{C}^2=(AC\times CP+AB\times BQ)$.

Answer 1

Given $\angle P=\angle Q={90}^{\circ }$

In $△ABP$

${AB}^2={AP}^2+{BP}^2$(pythagoras theorem)

In $△BPC$

${BC}^2={BP}^2+{PC}^2$(pythagoras theorem)

${BC}^2={AB}^2-{AP}^2+{(AP+AC)}^2$

${BC}^2={AB}^2+{AC}^2+2AP\times AC.....\left(2\right)$

In $△AQC$

${AC}^2={AQ}^2+{QC}^2$(pythagoras theorem)

In $△BQC$

${BC}^2={CQ}^2+{BQ}^2$ (pythagoras theorem)

${BC}^2={AC}^2-{AQ}^2+{(AB+AQ)}^2$

${BC}^2={AC}^2+{AB}^2+2ABAQ......\left(3\right)$

add (2) and (3)

$2{BC}^2=2{AC}^2+2{AB}^2+2AP\times AC+2AB\times AQ$

$2{BC}^2=2AC(AC+AP)+AB(AB+AQ)$

$2{BC}^2=2(AC\times PC)+2(AB\times BQ)$

${BC}^2=AC\times CP+AB\times BQ$

Hence proved