Proving Quadrilateral is a Parallelogram (II)

Question

Given:

1) Quadrilateral ABCD, all interior angles < 180°

2) AB ≅ CD

3) ∡B = ∡D > 90°

Prove:

ABCD is a parallelogram.

Answer 1

to prove the supposition,

absolutely need angle B+D>180 degrees

law of sines

sinB/AC=sin(DAC)/CD

sinD/AC=sin(BCA)/AB

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sin(DAC)=sin(BCA),,,assume sin^-1(DAC)<90 degrees (1st quadrant)

angle DAC=angle BCA and the quadrilateral is parallelogram

or angle

BCA =180- angle DAC and quadrilateral is not a parallelogram

because angles B+D>180 degrees,

angle BCA is precluded from being equal tp180- angle DAC

A+C<180 given

A=DAC+BAC

C=BCA+DCA

if BCA=180-DAC

A+C=DAC+BAC+(180-DAC)+DCA=180+BAC+DCA>180 contradiction

triangle ADC is congruent to ABC and ABCD is parallelogram

since both opposite sides are equal