Question

how to proceed next...how to take k as common and prove it...?

Answer 1

You have already solved it. Just one or two more steps.

$\frac{bc}{qr} = \frac{ \sqrt{ {k}^{2} {pq}^{2} -{k}^{2} {pr}^{2} } }{ \sqrt{ {pq}^{2} - {pr}^{2} } }$

$\frac{bc}{qr} = \frac{ \sqrt{ {k}^{2} ({pq}^{2} - {pr}^{2}) } }{ \sqrt{ {pq}^{2} - {pr}^{2} } }$

$\frac{bc}{qr} = \frac{k \sqrt{{pq}^{2} - {pr}^{2} } }{ \sqrt{ {pq}^{2} - {pr}^{2} } }$

$\frac{bc}{qr} = k$

Thus, both triangles are similar. (SSS similarity condition)

Thus, Angle B = Angle Q. ( corresponding angles)