4.2 The case of the Missing Diagram

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Transcript 4.2 The case of the Missing Diagram

Slide 1

Warm Up


Slide 2

Chapter 4.2
The Case of the Missing
Diagram


Slide 3

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


Slide 4

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.


Slide 5

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.


Slide 6

Now draw and label all you know.
You can label everything on the diagram
to help you make the proof.
A

Given:

ABC is isosceles

Base BC
AD is a median
B

D

Prove: BD

C

AD and bisects BC


Slide 7

NOTICE!!!

You can label everything on
a diagram to help you
make the proof.
Some problems you only have to draw,
label, write the givens and what to
prove.
Others you also have to prove.


Slide 8

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?


Slide 9

X

Given: Δ XYZ is equilateral

P

Z

R

Q

Y

PY, RZ and QX are medians
~
~
Prove: PY = RZ = QX


Slide 10

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.


Slide 11

C

T

R

A

Given: Δ CAT is isosceles, with base
TA.
CR is a median.
~ Δ ARC
Prove: Δ TRC =


Slide 12

Try this one!
If each pair of opposite sides of a foursided figure are congruent, then the
segments joining opposite vertices
bisect each other.
Draw

Write Given:
Write Prove:

Write proof!


Slide 13

A

D

B

E
C

Given:

~ CD
AB =
AD ~
= BC

Prove:

AC bisects BD

BD bisects AC


Slide 14

~ Δ CDA by SSS, and thus,
Δ ABC =
= Δ BAD ~
= Δ DCB by SSS, and thus,
= Thus Δ ABE ~
= Δ CDE by ASA, and then
AE ~
= EC and DE ~
= EB.
A

B

E
D

C