Question

3 in quadrilateral abcd on the right, given ab = dc and ad = bc, prove that quadrilateral abcd is a parallelogram as follows. fill the appropriate words in the blanks.
way of thinking
to prove that it is a parallelogram, use the definition that the two sets of opposite sides are parallel.
based on that thinking, show that the two triangles formed by drawing diagonal line ac are congruent, so that alternate angles are equal.
proof
draw diagonal ac. in △abc and △
given: ab = ……① bc = ……②
also is shared ……③
from ①, ②, and ③, since the three are equal respectively,
△ △
from ∠bac = ∠, since the alternate angles are equal, ab || dc.
from ∠bca = ∠, since the alternate angles are equal, ad || bc.
since pairs of sides are parallel, quadrilateral abcd is a parallelogram.
4 in right quadrilateral abcd, given ad || bc and ad = bc, prove that quadrilateral abcd is a parallelogram as follows. fill the appropriate words in the blanks.
proof
draw diagonal ac. in △abc and △,
given: = ……①
from ad || bc, ∠ = ∠ ……②
also, ac is shared ……③
from ①, ②, and ③, since the two and the included are equal respectively.
△ ≅ △.
since the measures of corresponding angles are equal in congruent figures, ∠ = ∠
since the alternate angles (∠bac and ∠dca) are equal, ab || dc.
from ab || dc and ad || bc,
since pairs of sides are parallel, quadrilateral abcd is a parallelogram.
★ like 3 and 4, the definition of a parallelogram can be derived from its properties.

Explanation:

Step1: Prove triangle congruence for question 3

• In $\triangle ABC$ and $\triangle CDA$ (diagonal $AC$ divides the quadrilateral).
• Given $AB = DC$ (①), $BC = AD$ (②), and $AC$ is shared (③).
• Since the three sides are equal respectively, by SSS (Side - Side - Side) congruence criterion, $\triangle ABC\cong\triangle CDA$.
• From $\angle BAC=\angle DCA$, so $AB\parallel DC$ (alternate interior angles are equal, lines are parallel).
• From $\angle BCA=\angle DAC$, so $AD\parallel BC$.
• Since two pairs of opposite sides are parallel, quadrilateral $ABCD$ is a parallelogram.

Step2: Prove triangle congruence for question 4

• Draw diagonal $AC$. In $\triangle ABC$ and $\triangle CDA$.
• Given $AD = BC$ (①).
• From $AD\parallel BC$, $\angle DAC=\angle BCA$ (②, alternate interior angles).
• Also, $AC$ is shared (③).
• Since two sides and the included angle are equal respectively, by SAS (Side - Angle - Side) congruence criterion, $\triangle ABC\cong\triangle CDA$.
• Since the measures of corresponding angles are equal in congruent figures, $\angle BAC=\angle DCA$.
• Since $AB\parallel DC$ and $AD\parallel BC$, and two pairs of opposite sides are parallel, quadrilateral $ABCD$ is a parallelogram.

Answer:

For question 3:

• First blank: $CDA$
• Second blank: $DC$
• Third blank: $AD$
• Fourth blank: $AC$
• Fifth blank: $sides$
• Sixth blank: $ABC$
• Seventh blank: $CDA$
• Eighth blank: $DCA$
• Ninth blank: $DAC$
• Tenth blank: $two$
• Eleventh blank: $opposite$

For question 4:

• First blank: $CDA$
• Second blank: $AD$
• Third blank: $BC$
• Fourth blank: $DAC$
• Fifth blank: $BCA$
• Sixth blank: $sides$
• Seventh blank: $angle$
• Eighth blank: $ABC$
• Ninth blank: $CDA$
• Tenth blank: $BAC$
• Eleventh blank: $DCA$
• Twelfth blank: $two$
• Thirteenth blank: $opposite$