Quadrilaterals Question

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Given that

$ABCD$ is a parallelogram and

$bar{BD}~=bar{AC}$ (line segments not pictured) prove that

$ABCD$ is a rectangle.

Rectangle ABCD

$\ \ \ \ \ \ \ \ \ \ \ \ " Statement " \ \ \ \ \ \ \ \ \ \ \ \$	$" Reason "$
$1. bar{AC} ~= bar{BD}$	$1. "Given"$
$2. ABCD " is a parallelogram"$	$2. "Given"$
$3. bar{AB} ~= bar{DC}$	$3. "Opposite sides of a parallelogram are congruent"$
$4. bar{AD} ~= bar{AD}$	$4. "Reflexive Property"$
$5.$	$5. "Side-Side-Side Postulate"$
$6. ang BAD ~= ang CDA$	$6. "Corresponding angles of congruent"$ $\ \ \ \ \ \ \ " triangles are congruent"$
$7. bar{AB} " \|\| " bar{CD}$	$7. "Opposite sides of a parallelogram are parallel"$
$8. m ang BAD + m ang CDA = 180°$	$8. ""$
$9. m ang BAD = m ang CDA$	$9. "Definition of congruent angles"$
$10. m ang BAD + m ang BAD = 180°$	$10. ""$
$11. 2m ang BAD = 180°$	$11. "Distributive Property"$
$12. m ang BAD = 90°$	$12. "Division Property of Equality"$
$13. m ang CDA = 90°$	$13. m ang CDA = m ang BAD$
$14.$	$14. "Definition of right angles"$
$15. bar{BC} " \|\| " bar{AD}$	$15. ""$
$16. m ang BAD + m ang ABC = 180°,$ $\ \ \ \ \ m ang ADC + m ang DCB = 180°$	$16. "Same side interior angles are supplementary"$
$17. 90° + m ang ABC = 180°,$ $\ \ \ \ \ 90° + m ang DCB = 180°$	$17. "Substitution Property of Equality"$
$18. m ang ABC = 90°,$ $\ \ \ \ \ m ang DCB = 90°$	$18. "Subtraction Property of Equality"$
$19. ang ABC, ang DCB " are right angles"$	$19. "Definition of right angles"$
$20. ABCD " is a rectangle"$	$20. ""$

Grade 10 Quadrilaterals CCSS: HSG-CO.C.11

What is the missing statement in step 5?

$Delta ABC ~= Delta DCB$
$Delta ABD ~= Delta DCA$
$Delta ADC ~= Delta ACB$
$Delta ABD ~= Delta DCB$

Question Info

Quadrilaterals Question

Grade 10 Quadrilaterals CCSS: HSG-CO.C.11