Question

in parallelogram abcd on the right, draw the two diagonal lines and one straight line passing through intersection point o of the diagonal lines in the direction of sides ad and bc. then, let the intersection points of sides ad and bc be e and f respectively. in this case, prove that eo = fo as follows.
(1) draw a line segment on the right figure according to the problem.
(2) fill in the blanks and complete the proof.
proof
in △aoe and △\boxed{cof},
since the diagonals intersect at the midpoint of each,
(the third property of a parallelogram)
\boxed{ao}=\boxed{oc}cdotscdots①
from ad || cb, ∠\boxed{oae}=∠\boxed{ocf}cdotscdots②
since the opposite angles are equal,
∠\boxed{aoe}=∠\boxed{cof}cdotscdots③
from ①, ②, and ③, since the two \boxed{angles} and the included \boxed{side} are equal respectively,
△\boxed{aoe}≌△\boxed{cof}.
therefore, \boxed{eo}=\boxed{fo}.

Explanation:

Step1: Identify the triangles

In parallelogram \(ABCD\), we consider \(\triangle AOE\) and \(\triangle COF\).

Step2: Use the property of parallelogram diagonals

Since the diagonals of a parallelogram bisect each other, \(AO = OC\).

Step3: Use the property of parallel lines

Because \(AD\parallel CB\), \(\angle OAE=\angle OCF\) (alternate - interior angles).

Step4: Use the property of vertical angles

\(\angle AOE=\angle COF\) (vertical angles are equal).

Step5: Apply the ASA (Angle - Side - Angle) congruence criterion

Since two angles (\(\angle OAE=\angle OCF\) and \(\angle AOE=\angle COF\)) and the included side (\(AO = OC\)) are equal, \(\triangle AOE\cong\triangle COF\) by ASA.

Step6: Use the property of congruent triangles

If \(\triangle AOE\cong\triangle COF\), then their corresponding parts are equal. So \(EO = FO\).

Answer:

In \(\triangle AOE\) and \(\triangle COF\), \(AO = OC\), \(\angle OAE=\angle OCF\), \(\angle AOE=\angle COF\), \(\triangle AOE\cong\triangle COF\) (ASA), and \(EO = FO\).