Question

In the figure given below , OL is perpendicular to AB and OM is perpendicular to BC such that OL = OM. Is $\Delta OLB\cong \Delta OMB$?

Answer 1

$Given,$

$OL=OM$

$\angle OLB$ = $\angle OMB,$ $and$ $they$ $are$ $right$ $angles.$

$So,$

$O{L}^2+B{L}^2=O{B}^2$

$O{M}^2+B{M}^2=O{B}^2$

$O{L}^2+B{L}^2$ = $O{M}^2+B{M}^2$

$B{L}^2$ $=$ $B{M}^2$

$\therefore$ $BL=BM$

$\angle OLB$ = $\angle OMB$

$OL=OM$

$By$ $SAS$ $congruence,$ $\Delta OLB\cong \Delta OMB$