Question

check again
●the five conditions of a parallelogram
① two pairs of opposite sides are equal respectively (the first property of a parallelogram)
② two pairs of opposite angles are equal respectively (the second property of a parallelogram)
③ two diagonals intersect at the midpoint of each (the third property of a parallelogram)
④ two pairs of opposite sides are parallel respectively (the definition of a parallelogram)
⑤ a pair of opposite sides are both parallel and equal in length
1 in parallelogram abcd on the right, given the midpoints of sides ab,
bc, cd, and da are e, f, g, and h respectively, prove that
quadrilateral efgh is a parallelogram as follows.
fill the appropriate words in the blanks.
(hint)
use the condition ① of a parallelogram.
(proof)
in △aeh and △
from ah = 1/2 ad, cf = 1/2 bc, and ad = bc,
ah = ……①
from ae = 1/2 ab, cg = 1/2 dc, and ab = dc,
ae = ……②
since opposite angles of a parallelogram are equal,
∠eah = ∠ ……③
from ①, ②, and ③, since two and the included are equal respectively,
△ ≅ △. therefore eh =
in the same way,
since △ebf ≅ △,
from eh = gf and ef = gh,
since pairs of sides are equal respectively,
quadrilateral efgh is a parallelogram.

Explanation:

Step1: Find the congruent triangle

In $\triangle AEH$ and $\triangle CGF$.

Step2: Calculate the length of $AH$

From $AH=\frac{1}{2}AD$, $CF = \frac{1}{2}BC$, and $AD = BC$, $AH = CF$.

Step3: Calculate the length of $AE$

From $AE=\frac{1}{2}AB$, $CG=\frac{1}{2}DC$, and $AB = DC$, $AE = CG$.

Step4: Find the equal angle

Since opposite angles of a parallelogram are equal, $\angle EAH=\angle FCG$.

Step5: Prove the congruence of triangles

From ①, ②, and ③, since two sides and the included angle are equal respectively, $\triangle AEH\cong\triangle CGF$. Therefore $EH = GF$.

Step6: Prove another pair of congruent triangles

In the same way, since $\triangle EBF\cong\triangle GDH$, $EF = GH$.

Step7: Prove the quadrilateral is a parallelogram

From $EH = GF$ and $EF = GH$, since two pairs of opposite sides are equal respectively, quadrilateral $EFGH$ is a parallelogram.

Answer:

In $\triangle AEH$ and $\triangle CGF$. $AH = CF$. $AE = CG$. $\angle EAH=\angle FCG$. sides, angle. $\triangle AEH\cong\triangle CGF$. $EH = GF$. $\triangle GDH$, $EF = GH$. two, opposite.