Ques.No.1
In #DeltasABO# and #COD#
#/_AOB=/_COD# - vertically opposite angles
#/_ABO=/_DCO# - alternate opposite angles as #AB#||#CD#
#/_BAO=/_ODC# - alternate opposite angles as #AB#||#CD#
As such #DeltasABO# and #COD# are similar and hence
#(CO)/(OB)=(CD)/(AB)# i.e. #x/2=12/4#
and #x=12/4xx2=6# #cm.#
Ques.No.2
As #AB#||#CE# and #AC# is transverse #/_a=70^@# as they are alternate interior angles.
Further as #AB#||#CE# and #BC# is transverse #/_b=/_c# as they are corresponding angles.
Now in #DeltaABC#, we have #AB=AC#, hence #/_ACB=/_ABC=b^@#
Hence #2b^@+a^@=2b^@+70^@=180^@#
or #2b^@=180^@-70^@=110^@# i.e. #b=55#
and as #/_c=/_b=55^@#, #c=55#