2 in parallelogram abcd on the right, given ae and cf are the perpendicular lines drawn from vertices a and c on diagonal bd respectively, answer the questions below.
(1) draw a line segment on the right figure according to the question.
(2) prove that quadrilateral aecf formed by connecting points a with e, and points c with f respectively is a parallelogram. fill the appropriate words in the blanks.
proof
in $ \triangle aeb $ and $ \triangle \square $,
from the property of a parallelogram,
$ \square = \square \cdots \cdots $ ①
given: $ \angle \square = \angle \square = 90 ^ { \circ } \cdots \cdots $ ②
from $ a b \parallel c d , \angle \square = \angle \square \cdots \cdots $ ③
from ①, ②, and ③, since the $ \square $ and one $ \square $ are equal respectively,
$ \triangle \square \cong \triangle \square $. therefore, $ a e = c f $.
from $ a e \perp b d $ and $ c f \perp b d , a e \parallel \square $.
from $ a e \parallel c f $ and $ a e = c f $,
since a pair of sides are both $ \square $ and $ \square $,
quadrilateral aecf is a parallelogram.
3 in parallelogram abcd on the right, let o be the intersection point of two diagonals ac and bd. then, place two points e and f on diagonal bd, so that $ e o = f o $. in this case, prove that quadrilateral aecf is a parallelogram as follows. fill the appropriate words in the blanks.
proof
since point o is the intersection point of the diagonal lines of parallelogram abcd,
$ a o = \square \cdots \cdots $ ①
given: $ \square = \square \cdots \cdots $ ②
from ① and ②,
since two $ \square $ intersect at the $ \square $ of each,
quadrilateral aecf is a parallelogram.

Question

Accepted Answer

# Explanation:
## Step1: Analyze the first proof (Problem 2)
- In $\triangle AEB$ and $\triangle CFD$, from the property of a parallelogram $AB = CD$ (opposite sides of a parallelogram are equal).
- Given $\angle AEB=\angle CFD = 90^{\circ}$ (AE and CF are perpendicular to BD).
- From $AB\parallel CD$, $\angle ABE=\angle CDF$ (alternate - interior angles).
- Since two angles and one side (AAS - Angle - Angle - Side) are equal respectively, $\triangle AEB\cong\triangle CFD$.
- From $AE\perp BD$ and $CF\perp BD$, $AE\parallel CF$.
- Since a pair of sides (AE and CF) are both parallel and equal, quadrilateral $AECF$ is a parallelogram.

## Step2: Analyze the second proof (Problem 3)
- Since point $O$ is the intersection point of the diagonal lines of parallelogram $ABCD$, $AO = CO$ (diagonals of a parallelogram bisect each other).
- Given $EO = FO$.
- From $AO = CO$ and $EO = FO$, since two diagonals ($AC$ and $EF$) intersect at the mid - point of each, quadrilateral $AECF$ is a parallelogram (a quadrilateral whose diagonals bisect each other is a parallelogram).

# Answer:
For problem 2:
- First blank: $CFD$
- Second and third blanks: $AB$, $CD$
- Fourth and fifth blanks: $\angle AEB$, $\angle CFD$
- Sixth and seventh blanks: $\angle ABE$, $\angle CDF$
- Eighth blank: two angles
- Ninth blank: side
- Tenth and eleventh blanks: $AEB$, $CFD$
- Twelfth blank: $CF$
- Thirteenth and fourteenth blanks: parallel, equal

For problem 3:
- First blank: $CO$
- Second and third blanks: $EO$, $FO$
- Fourth blank: diagonals
- Fifth blank: mid - point

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the first proof (Problem 2)

Step2: Analyze the second proof (Problem 3)

Answer: