Warm Up. Chapter 4.2 The Case of the Missing Diagram

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ChapteChapter 4.2r 4.2The Case of The Case of the Missing the Missing

DiagramDiagram

Organize the information in, Organize the information in, and draw diagrams for, and draw diagrams for, problems presented in problems presented in

words.words.

Set up this problem:

An isosceles triangle and the median to the base.

Draw the shape, label everything

Write the givens and what you want to prove.

Given: an isosceles triangle and the median to the base.

Prove: The median is the perpendicular bisector of the base.

Notice: There are two conclusions to made:

1.The median is perpendicular to the base.

2. The median bisects the base.

Now draw and label all you know.

You can label everything on the diagram to help you make the proof.

A

B CD

Given: ABC is isosceles

Base BC

AD is a median

Prove: BD AD and bisects BC

NOTICE!!!NOTICE!!!

You can label everything on You can label everything on a diagram to help you a diagram to help you make the proof.make the proof.

Some problems you only have to draw, label, write the givens and what to prove.

Others you also have to prove.

Remember If….then….

Sometimes you will see these in reverse.

The medians of a triangle are congruent if the triangle is equilateral.

Draw and set up the proof.

Write down the givens you need.

What do you need to prove?

Given: Δ XYZ is equilateral

PY, RZ and QX are medians

Prove: PY = RZ = QX~ ~

X

YZ

PR

Q

The median to the base of an isosceles triangle divides the triangle into two congruent triangles.

Draw, write givens and what to prove, then prove.

C

ATR

Given: Δ CAT is isosceles, with base TA.

CR is a median.

Prove: Δ TRC = Δ ARC~

Try this one!

If each pair of opposite sides of a four-sided figure are congruent, then the segments joining opposite vertices bisect each other.

Draw

Write Given:

Write Prove:

Write proof!

A B

CD

E

Given: AB = CD

AD = BC

Prove: AC bisects BD

BD bisects AC

~

~

Δ ABC = Δ CDA by SSS, and thus, <BAC = <DCA.

Δ BAD = Δ DCB by SSS, and thus, <ABD = <CDB.

Thus Δ ABE = Δ CDE by ASA, and then AE = EC and DE = EB.

~~

~~

~~ ~

A B

CD

E