ABCD is a cyclic quadrilateral. Diagonals BD and AC intersect each other at E. If ∠BEC = 138° and ∠ECD = 35°, then what is the measure of ∠BAC ?

103°

\(\angle\)BEC and \(\angle\)CED are on the same straight lines

\(\angle\)BEC = 138

\(\angle\)CED = 180 - 138

= \(\angle\)CED = \({42}^\circ\)

In \(\Delta \)CDE, \(\angle\)CED = 42 and \(\angle\)DCE = 35

\(\angle\)CDE = 180 - (42 + 35)

= \(\angle\)CDE = \({103}^\circ\)

\(\angle\)BAC and \(\angle\)BDC are on the same arc BC

We know that in cyclic quadrilateral angles on the same arc are always same.

\(\angle\)BAC = \({103}^\circ\)

Therefore, \(\angle\)BAC is \({103}^\circ\).