Question

In ΔABC, ∠ACB is obtuse angle, seg AD ⊥ seg BC. Prove that:
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${AB}^2$ = <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${BC}^2$ + <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${AC}^2$ + 2BC × CD

[3 Marks]

Answer 1

In the figure, seg AD ⊥ seg BC
Let AD = p, AC = b, AB = c,
BC = a
and DC = x.
DB = a + x [1 Mark]

In <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Δ ADB, by Pythagoras theorem,
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $c^2$ = ${(a+x)}^2$ + $p^2$
$c^2$ = $a^2$ + 2ax + $x^2$ + $p^2$ .......... (I)
Similarly, in <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Δ ADC
$b^2$= $x^2$ + $p^2$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ∴ $p^2$ = $b^2$ - $x^2$.......... (II) [1 Mark]

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ∴ substituting the value of $p^2$ from (II) in (I)
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ∴ $c^2$ = $a^2$ + 2ax + $b^2$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ∴ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${AB}^2$ = <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${BC}^2$ + ${AC}^2$ + 2BC × CD. [1 Mark]