Question

in the figure,
$segPQ\parallel sideBC$ and
$segQR\parallel sideAB$.
i) Find $\frac{AQ}{QC}$.
ii) What would be $\frac{CR}{RB}$? Hence find $\frac{BR}{RC}$

Answer 1

In $\Delta APQ$ & $\Delta ABC$

$\angle BAC=\angle PAQ$ (same angle)

$\angle ABC=\angle APQ$ ($PQ\parallel BC$)

$\angle AQP=\angle ACB$ ($PQ\parallel BC$)

Triangles are similar

$\frac{AC}{AQ}=\frac{AB}{AP}=\frac{5}{3}$

$\frac{AC}{AQ}-1=\frac{AC-AQ}{AQ}=\frac{5}{3}-1=\frac{2}{3}$

$\Rightarrow \frac{QC}{AQ}=\frac{2}{3}$ , $\frac{AQ}{QC}=\frac{3}{2}$

(ii) $\Delta CQR$ and $\Delta CAB$

$\angle ACB=\angle QCR$ (same angle)

$\angle CQR=\angle CAB$ $(QR\parallel AB)$

$\angle CRQ=\angle CBA$ $(QR\parallel AB)$

Triangles are similar

$\frac{RB}{CR}+1=\frac{RB+CR}{CR}=\frac{CB}{CR}$

$\frac{CB}{CR}=\frac{CA}{CQ}=\frac{5}{2}$

$\frac{CB}{CR}=\frac{5}{2}$

$\frac{RB}{CR}=\frac{5}{2}-1$

$\frac{RB}{CR}=\frac{3}{2}$