Question

In the figure (1) given below, AB | CR and LM | QR
Prove that $\frac{BM}{MC}=\frac{AL}{LQ}$

Answer 1

$In△MBA\&△MCR$

$\angle MBA=\angle MCR${ given $AB||CR$ and we take $BC$ as transversal line}

$\angle AMB=\angle RMC$ { vertically opposite angle}

$So,△MBA\sim △MCR${ By AA rule}

$Then\frac{BM}{MC}=\frac{AM}{MR}-\left(i\right)$

and $In△ARQ$,given $LM||QR$,so from $B.P.T$ we got:

$\frac{AM}{MR}=\frac{AL}{LQ}-\left(ii\right)⟹\frac{BM}{MC}=\frac{AL}{LQ}$