Question

Given: start overline, D, E, end overline, is perpendicular to, start overline, B, C, end overline, comma
DE
⊥
BC
, start overline, B, F, end overline, is perpendicular to, start overline, D, C, end overline
BF
⊥
DC
and A, B, C, DABCD is a rhombus.

Prove: angle, C, D, E, \cong, angle, C, B, F∠CDE≅∠CBF.
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
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

A, B, C, DABCD is a rhombus
Given
2
angle, D, E, C∠DEC is a right angle _

Answer 1

Fill the dropdowns with these reasons for the successive statements:

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

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

4. ∠DCE ≅ ∠BCF — Reason: They are the same angle at C (reflexive / common angle).

5. ΔCDE ∼ ΔCBF — Reason: AA (two angles of one triangle are congruent to two angles of the other: ∠DEC ≅ ∠CFB and ∠DCE ≅ ∠BCF).

6. ∠CDE ≅ ∠CBF — Reason: Corresponding angles of similar triangles are congruent.

This proves ∠CDE ≅ ∠CBF.