Question

Given: A, B, C, DABCD is a parallelogram, start overline, F, C, end overline, \cong, start overline, E, C, end overline, comma
FC
≅
EC
, start overline, D, E, end overline, is perpendicular to, start overline, B, C, end overline
DE
⊥
BC
and start overline, B, F, end overline, is perpendicular to, start overline, D, C, end overline, .
BF
⊥
DC
.

Prove: A, B, C, DABCD is a rhombus.
In the proof shown in the table below, the statements and the first reason are filled in for you. Use the dropdown menu in the Reason column to select the reason for the next statement. Once you have selected a reason, a new statement will appear in a new row below. This will continue until you have selected the reason for the final statement, and then the answer area will appear with a submit button. The diagram associated with this problem is located at the bottom of the page, below the proof, and is available as a tactile printout. There may be a clarifying note below the diagram.
Step Statement Reason
1
A, B, C, DABCD is a parallelogram
start overline, F, C, end overline, \cong, start overline, E, C, end overline
FC
≅
EC

start overline, D, E, end overline, is perpendicular to, start overline, B, C, end overline
DE
⊥
BC

start overline, B, F, end overline, is perpendicular to, start overline, D, C, end overline
BF
⊥
DC

Given
2
angle, D, E, C∠DEC is a right angle _

Answer 1

Fill in the missing reasons as follows and the proof is complete.

1. ABCD is a parallelogram; FC ≅ EC; DE ⟂ BC; BF ⟂ DC. (Given)

2. ∠DEC is a right angle. — Reason: Perpendicular lines form right angles (DE ⟂ BC and EC lies on BC).

3. ∠CFB is a right angle. — Reason: Perpendicular lines form right angles (BF ⟂ DC and FC lies on DC).

4. ΔDEC ≅ ΔCFB. — Reason: Hypotenuse-Leg (HL) Congruence for right triangles (each triangle is right and EC ≅ FC is a leg).

5. DC ≅ CB. — Reason: Corresponding parts of congruent triangles are congruent (CPCTC).

6. AB ≅ BC. — Reason: Opposite sides of a parallelogram are congruent, so AB ≅ CD and with CD ≅ CB from step 5 we get AB ≅ BC.

7. Therefore ABCD is a rhombus. — Reason: A parallelogram with a pair of adjacent congruent sides is a rhombus (all four sides equal).

Thus ABCD is a rhombus.