Problem
MO is perpendicular to CD, so angle MOC is 90 deg on the left, split by OE which sits 70 deg from MO.
Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.
4.G.A.2Draw A DiagramOE splits the 90 deg angle MOC into 70 deg and angle EOC, so EOC = 90 − 70 = 20 deg.
The upward segment to OE uses 70 deg of the 90 deg quarter-turn, leaving 20 deg from OE down to the horizontal line OC.
4.MD.C.7Identify SubproblemsOE/OF and OC/OD are opposite ray pairs, so angle DOF (which is a) is EOC's vertical angle — equal to it.
Crossing lines make equal angles straight across, so the small 20 deg angle on the upper-left reappears as a on the lower-right.
4.MD.C.7Identify SubproblemsAngle a (angle DOF) is equal to angle EOC, the angle sitting across the crossing on the upper left.
Angle DOF and angle EOC are the two angles that face each other where the straight lines CD and EF cross, and such opposite angles always have the same size.
Both angle EOC and angle DOF lean on the very same neighbor, angle EOD, and each of them teams up with that neighbor along a straight line to reach 180 degrees.
Angle EOC and angle EOD sit side by side along the straight line CD, so together they fill one straight angle.
Angle EOD and angle DOF sit side by side along the straight line EF, so together they fill one straight angle.
Angle EOC is 180 minus angle EOD, and angle DOF is also 180 minus angle EOD, so both equal the same amount and therefore equal each other.
Take 70 deg out of the 90 deg right angle to get 20 deg, then bounce it straight across the crossing: angle a is 20 deg!