Parallelogram w/ Congruent Diagonals is Rectangle (Grade 10)

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Rectangle ABCDFill in the reasons for the following proof.

Given that [math]ABCD[/math] is a parallelogram and [math]bar{BD}~=bar{AC}[/math] (line segments not pictured) prove that [math]ABCD[/math] is a rectangle.
[math] \ \ \ \ \ \ \ \ \ \ \ \ " Statement " \ \ \ \ \ \ \ \ \ \ \ \ [/math][math] " Reason "[/math]
[math]1. bar{AC} ~= bar{BD} [/math][math]1. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [/math]
[math]2. ABCD " is a parallelogram"[/math][math]2. ""[/math]
[math]3. bar{AB} ~= bar{DC} [/math][math]3. ""[/math]
[math]4. bar{AD} ~= bar{AD} [/math][math]4. ""[/math]
[math]5. Delta ABD ~= Delta DCA [/math][math]5. ""[/math]
[math]6. ang BAD ~= ang CDA[/math][math]6. [/math]
[math]7. bar{AB} " || " bar{CD} [/math][math]7. ""[/math]
[math]8. m ang BAD + m ang CDA = 180°[/math][math]8. ""[/math]
[math]9. m ang BAD = m ang CDA [/math][math]9. ""[/math]
[math]10. m ang BAD + m ang BAD = 180° [/math][math]10. ""[/math]
[math]11. 2m ang BAD = 180° [/math][math]11. ""[/math]
[math]12. m ang BAD = 90° [/math][math]12. ""[/math]
[math]13. m ang CDA = 90° [/math][math]13. [/math]
[math]14. ang BAD and ang CDA " are right angles" [/math][math]14. [/math]
[math]15. bar{BC} " || " bar{AD} [/math][math]15. ""[/math]
[math]16. m ang BAD + m ang ABC = 180°, [/math] [math] \ \ \ \ \ m ang ADC + m ang DCB = 180° [/math][math]16. ""[/math]
[math]17. 90° + m ang ABC = 180°, [/math] [math] \ \ \ \ \ 90° + m ang DCB = 180° [/math][math]17. ""[/math]
[math]18. m ang ABC = 90°, [/math] [math] \ \ \ \ \ m ang DCB = 90° [/math][math]18. ""[/math]
[math]19. ang ABC, ang DCB " are right angles" [/math][math]19. ""[/math]
[math]20. ABCD " is a rectangle" [/math][math]20. ""[/math]



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