Question
In $\triangle\text{ABC},\ \angle\text{A}$ is obtuse, $\text{PB}\perp\text{AC}$ and $\text{QC}\perp\text{AB}$ Prove that:
$BC^2 = (AC \times CP + AB \times BQ)$
$BC^2 = (AC \times CP + AB \times BQ)$
✓
Answer
In $\triangle\text{BPC},$ by pythagoras theorem
$BC^2 = BP^2 + PC^2$
$\Rightarrow BC^2 = AB^2 - AP^2 + (AP + AC)^2 [$By pythagoras theorem$]$
$\Rightarrow BC^2 = AB^2 + AC^2 + 2AP \times AC .....(ii)$
In $\triangle\text{BQC},$ by pythagoras theorem,
$BC^2 = CQ^2 + BQ^2$
$\Rightarrow BC^2 = AC^2 - AQ^2 + (AB + AQ)^2 [$By pythagoras theorem$]$
$\Rightarrow BC^2 = AC^2 - AQ^2 + AB^2 + AQ^2 + 2AB \times AQ$
$\Rightarrow BC^2 = AC^2 + AB^2 + 2AB \times AQ ....(iii)$
Add equations $ (ii)\ \&\ (iii)$
$2BC^2 = 2AC^2 + 2AB^2 + 2AP \times AC + 2AB \times AQ$
$\Rightarrow 2BC^2 = 2AC^2 + 2AB^2 + 2AP \times AC + 2AB \times AQ$
$\Rightarrow 2BC^2 = 2AC[AC + AP] + AB[AB + AQ]$
$\Rightarrow 2BC^2 = 2AC \times PC + 2AB \times BQ$
$\Rightarrow BC^2 = AC \times PC + AB \times BQ [$Divide by $2]$
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