Question

In the given figure, DE || BC and CD || EF. Prove that ${AD}^2$ = AB $\times$ AF.


[3 Marks]

Answer 1

In ΔABC, we have
DE || BC
$\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{BD}{AD}$ = $\frac{CE}{AE}$ [Basic Proportionality Theorem]

$\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{BD}{AD}$ + 1 = $\frac{CE}{AE}$ + 1

$\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{BD+AD}{AD}$ = $\frac{CE+AE}{AE}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{AB}{AD}$ = $\frac{AC}{AE}$....(i)
[1 Mark]

In ΔADC, we have
FE || DC
$\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{DF}{AF}$ = $\frac{EC}{AE}$ [Basic Proportionality Theorem]

$\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{DF}{AF}$ + 1 = $\frac{EC}{AE}$ + 1

$\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{DF+AF}{AF}$ = $\frac{EC+AE}{AE}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{AD}{AF}$ = $\frac{AC}{AE}$......(ii)
[1 Mark]

From (i) and (ii), we get
$\frac{AB}{AD}$ = $\frac{AD}{AF}$

$\Rightarrow$ <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${AD}^2$ = AB $\times$ AF. [1 Mark]