Question

Given: A, B, C, DABCD is a parallelogram and triangle, A, B, C, \cong, triangle, D, C, B, .△ABC≅△DCB.

Prove: start overline, A, B, end overline, is perpendicular to, start overline, B, C, end overline
AB
⊥
BC
.
In the table below, the first step of the proof is completed for you. To complete each following step, first use the dropdown menu in the "Statement" column to select the format of a statement. Once you select a format, a statement will appear with input boxes and/or dropdowns to complete, and a dropdown menu will appear in the reason column. Once you complete these inputs and dropdowns, navigate back to the first column and press the "try" button. If your statement and reason are valid, a new row will appear below. This process will continue until you have filled in a statement and reason for what you were asked to prove, and then the "try" button will cause the answer area to appear with a submit button. The diagram associated with this problem is located below the proof table, 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
triangle, A, B, C, \cong, triangle, D, C, B△ABC≅△DCB
Given
2
start overline, A, C, end overline, \cong, start overline, B, D, end overline
AC
≅
BD

Corresponding Parts of Congruent Triangles are Congruent (CPCTC). What are the remaining steps, statements, and reasons for this question?

Answer 1

Fill in the missing steps as follows.

1. ΔABC ≅ ΔDCB. — Given
2. AC ≅ BD. — Corresponding parts of congruent triangles are congruent (CPCTC).
3. Let E = AC ∩ BD. — Definition (introduce intersection of diagonals).
4. AE ≅ EC and BE ≅ ED. — Diagonals of a parallelogram bisect each other.
5. AE ≅ BE (hence AE ≅ BE ≅ CE ≅ DE). — From AC ≅ BD and the fact the diagonals are bisected, the halves are congruent (if AC = BD and AC = 2·AE, BD = 2·BE then AE = BE; together with AE = EC and BE = ED gives all four halves equal).
6. A, B, C, D are concyclic (E is equidistant from A, B, C, D). — If four points are all the same distance from a point, they lie on a circle (definition / locus of points equidistant from a center).
7. ∠A + ∠C = 180°. — Opposite angles of a cyclic quadrilateral are supplementary.
8. ∠A = ∠C. — Opposite angles of a parallelogram are congruent.
9. ∠A = 90°. — From ∠A + ∠C = 180° and ∠A = ∠C we get 2∠A = 180°, so ∠A = 90°.
10. ∠B = 90° (so AB ⟂ BC). — Adjacent angles in a parallelogram are supplementary, so if ∠A = 90° then ∠B = 90°, hence AB ⟂ BC.

Thus AB is perpendicular to BC.