Proving Concave Quadrilateral is a Parallelogram

Question

Given quadrilateral ABCD

with all interior angles less than 180 degrees,

and given that:

AB is congruent to CD

Angle B is congruent to angle D

Prove ABCD is a parallelogram

Answer 1

I made a couple of modifications to my previous answer. These are highlighted in red.

sin(B)/AC = sin(δ)/AB and sin(D)/AC = sin(θ)/CD (Law of Sines)

So sin(B)/AC =sin(D)/AC (since ∠B ≅ ∠D)

which means that sin(δ)/AB = sin(θ)/CD (substitution; they're equal to equal things)

So sin(δ) = sin(θ) (since AB = CD)

which means that either δ = θ or δ = 180 – θ

So ∠DAC and ∠BCA can be, but don't have to be, congruent angles

In the case where ∠DAC ≅ ∠BCA we have

ΔABC ≅ ΔCDA (AAS congruence)

segment BC ≅ segment DA (congruent parts of congruent triangles are congruent)

ABCD is a parallelogram → If both pairs of opposite sides of a quadrilateral are

congruent, then the quadrilateral is a parallelogram.

It is possible, however, to draw a quadrilateral that is not a parallelogram where

AB is congruent to CD, and

Angle B is congruent to angle D

A Desmos graph showing and example of this: desmos.com/calculator/hudpikaz0i

Answer 2

This was my own question, and through much thinking

and with help from William C.'s comments I have

found that the conclusion cannot be proved.

Given ABCD is a parallelogram, then it will have

as part of it's description the one side will be congruent

to its opposite side, and one angle congruent to

its opposite angle.

However just being given a concave (normal looking)

quadrilateral ABCD with AB congruent to CD

and angle B congruent to angle D (see diagram),

then it cannot be concluded that ABCD is a parallelogram.

Example:AB=CD=4√2,,,,AD=5,,,,BC=3,,,diagonal AC=√17

angle B=qngle D=45 degrees,,,

angle BCA=104 degrees,,,,,,angleDCA=59 degrees

so angle C=163

angle DAC=76 degrees, angle CAB=31 degrees

so angle A=107 degrees

Answer 3

Here are some facts with which to do the proof

∠A ≅ ∠B

∠A is supplementary to ∠D

AB || DC