The Diagonals of a Rectangle

02 Dec

CCSS-M.G-CO.C.11.Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.

How can you use rigid motions to show that the diagonals of a rectangle are congruent?

I had a few students come in during zero block to work on proofs. (If you read my last post on the diagonals of an isosceles trapezoid, you’ll know why.)

We were trying to show that AC=BD. Students used look for and make use of structure to compose the rectangle into two triangles. Which two triangles should we show congruent if we want to show that the diagonals are congruent?

∆ACD and ∆BDC.

One student showed the two triangles congruent by SAS. (Opposite sides of a rectangle are congruent, all angles in a rectangle are right and thus congruent, and CD=CD by reflexive.)

But a pair of girls wanted to use a rigid motion to show that the triangles were congruent.

First up: rotating.

Rotating which triangle?

∆ACD

The center of the rectangle. Where the diagonals meet.

How many degrees?

90˚

(I’m not sure whether they really thought we should rotate by 90˚ or they chose 90˚ because we seem to rotate by 90˚ and 180˚ more than any other angle measure.)

We need paper. And scissors.

What is the image of ∆ACD when we rotate it 90˚ about the intersection of the diagonals?

Not ∆BDC.

What is the image of ∆ACD when we rotate it 180˚ about the intersection of the diagonals?

∆ABC

How can we show that ∆ACD and ∆BDC are congruent?

A reflection?

About the perpendicular bisector of segments AB and CD.

Are you sure?

Yes.

Once the triangles are congruent, then the corresponding parts are congruent, and so we can conclude that the diagonals are congruent.

And so the journey continues … with an apology to my former students for not using scissors more often.