One theorem says that if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. I want to prove this using a two column format. Are my statements and reasons below correct?
Suppose we have parallelogram ABCD, where A is the bottom left vertex, B is the upper left vertex, C is the upper right vertex, and D is the bottom right vertex.
Statement/Reason
1. ABCD is a parallelogram/Given
2. ∠C is congruent to ∠A = 90°/Definition of parallelogram
3. m∠B is congruent to m∠ A = 90°/AD and BC are parallel lines cut by transversal AB, so corresponding angles A and B are congruent.
4. m∠B = m∠D = 90°/Definition of parallelogram
5. ABCD is a rectangle/ABCD has all right angles.
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Answers & Comments
Verified answer
Statement/Reason
1. ABCD is a parallelogram/Given
1A. ANGLE A IS A RIGHT ANGLE / GIVEN
1B. ANGLE A = 90 DEGREES / DEF OF RIGHT ANGLE
2. ∠C is congruent to ∠A = 90°/Definition of parallelogramNO - DEF OF PARA SAY BOTH PAIRS OF OPP SIDES ARE PARALLEL.......SHOULD BE ,,,,OPP. ANGLES OF A PARA ARE EQUAL
3. m∠B is congruent to m∠ A = 90°/AD and BC are parallel lines cut by transversal AB, so corresponding angles A and B are congruent.CHANGE 3. COMPLETELY
3. ANGLE B AND ANGLE A ARE SUPPLEMENTARY ANGLES, / iF LINES PARALLEL SAME SIDE INTERION ANGLES ARE SUPP.
3A. ANG A + ANG B = 180 DEGREES / DEF OF SUPP ANGLES
3B. 90 + ANG B = 180 / SUBSTITUTION
3C. ANG B = 90 / SUBTRACTION PROPERTY OF EQUALITY
4. m∠B = m∠D = 90°/Definition of parallelogramNO......OPP ANGLES OF A PARA ARE EQUAL
5. ABCD is a rectangle/ABCD has all right angles. DEF OF A RECTANGLE.....A PARA WITH ALL RIGHT ANGLES IS A RECTANGLE
HOPE YOU UNDERSTAND.......IF NOT READ BELOW IF ANY QUESTIONS
Mathcerely,
Robert Jones.......Master in Mathematics (Un of Notre Dame)
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"Suppose we have parallelogram ABCD, where A is the bottom left vertex, B is the upper left vertex, C is the upper right vertex, and D is the bottom right vertex."
You need to include the supposition that angle A is a right angle. You use that in you proof, but never explicitly stated it.
"3. m∠B is congruent to m∠ A = 90°/AD and BC are parallel lines cut by transversal AB, so corresponding angles A and B are congruent."
Angles A and B are not corresponding angles; they are consecutive interior angles aka same side interior angles. All you know about them is that A + B = 180 degrees. That will still get you to B=90 degrees, but you need more words.
I think that'll work, but I'd add another step because the corresponding angle to A isn't inside the parallelogram. So I'd say that ∠A is congruent to its correspondent ∠B ext. (the exterior angle to B). Then this angle lies on the straight line AB along with ∠B (the one inside the parallelogram) so they're supplementary. But ∠A = 90º and so ∠B = 90º as well so both those angles are right.
It's a small thing, but I believe it's required.
Sorry I couldn't be more formal, but it's been a while since I've had a formal geometry course.
I did the comparable actual concern...there replaced into greater efficient than one as end results of the coach it. additionally, you're able to digital mail your geometry instructor and ask her- she have been given a examine out the attempt so she'll understand what you're speaking approximately. i replaced into freaking out, because of the fact as quickly as I have been given out of the finding out room, each physique replaced into speaking approximately proving 2 triangles congruent and using cpctc yet I proved the angles congruent via substitute indoors angles.