Question

1 in parallelogram abcd on the right, given e and f are the midpoints of sides ab and cd respectively, prove that af = ce as follows. fill the appropriate words in the blanks. proof in △afd and △, since the opposite angles of a parallelogram are equal, (the second property of a parallelogram) ∠adf = ∠………① given: ad =………② be = 1/2 and df = 1/2. from ab = cd, be =………③ from ①, ②, and ③, since the two and the included are equal respectively, △≌△. since the sides of congruent figures are equal, af =.

Explanation:

Step1: Identify the congruent triangles

In parallelogram \(ABCD\), to prove \(AF = CE\), we consider \(\triangle AFD\) and \(\triangle CEB\).

Step2: Use the properties of parallelogram for angles

Since \(ABCD\) is a parallelogram, \(\angle ADF=\angle CBE\) (opposite angles of a parallelogram are equal).

Step3: Use the properties of parallelogram for sides

In parallelogram \(ABCD\), \(AD = BC\) (opposite sides of a parallelogram are equal). Also, \(BE=\frac{1}{2}AB\) and \(DF = \frac{1}{2}CD\), and since \(AB = CD\) (opposite sides of a parallelogram), then \(BE=DF\).

Step4: Apply the SAS (Side - Angle - Side) congruence criterion

From \(\angle ADF=\angle CBE\) (\(AD = BC\) and \(DF = BE\)), by the SAS congruence criterion, \(\triangle AFD\cong\triangle CEB\).

Step5: Conclude the equality of \(AF\) and \(CE\)

Since \(\triangle AFD\cong\triangle CEB\), and the sides of congruent figures are equal, \(AF = CE\).

Answer:

In \(\triangle AFD\) and \(\triangle CEB\), \(\angle ADF=\angle CBE\), \(AD = BC\), \(BE=\frac{1}{2}AB\), \(DF=\frac{1}{2}CD\), \(BE = DF\), since the two sides and the included angle are equal respectively, \(\triangle AFD\cong\triangle CEB\), \(AF = CE\).
So the blanks are filled as: \(CEB\), \(CBE\), \(BC\), \(AB\), \(CD\), \(DF\), sides, angle, \(AFD\), \(CEB\), \(CE\)