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The Case of the Missing Diagram

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Chapter 4.2 The Case of the Missing Diagram Objective: After studying this section, you will be able to organize the information in, and draw diagrams for, problems ... – PowerPoint PPT presentation

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Title: The Case of the Missing Diagram


1
Chapter 4.2
  • The Case of the Missing Diagram

2
  • Objective After studying this section, you will
    be able to organize the information in, and draw
    diagrams for, problems presented in words.

3
Set up a proof of the statement If two
altitudes of a triangle are congruent, then the
triangle is isosceles.
  • Draw the shape, label everything.
  • The if part of the statement is the given.
  • The then part of the statement is the prove.
  • Write the givens and what you want to prove.

4
A
  1. Draw a diagram to show two altitudes in a
    triangle. Label everything.
  2. Write your given statements.
  3. Write your prove statement.

E
B
D
C
Given BD CE are altitudes to AC AD of
ACD. BD ? CE. Prove ACD is isosceles.
5
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.
6
Remember If.then. Sometimes you will see these
in reverse. The medians of a triangle are
congruent if the triangle is equilateral.
  1. Draw the diagram.
  2. Write down the givens you need.
  3. What do you need to prove?

7
X
  1. Draw a diagram to show the medians in an
    equilateral triangle. Label everything.
  2. Write your given statements.
  3. Write your prove statement.

R
P
Y
Z
Q
Given XYZ is equilateral.PZ, RY, and QX are
medians. Prove PZ ? RY ? QX
8
One more time. 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.
  1. Draw
  2. Write Given
  3. Write Prove
  4. Write proof

9
A
B
E
C
D
Given AB ? CD AD ? BC Prove AC bisects
BD BD bisects AC
10
? ABC ? ? CDA by SSS, and thus, ltBAC ? ltDCA. ?
BAD ? ? DCB by SSS, and thus, ltABD ? ltCDB. Thus
? ABE ? ? CDE by ASA, and then AE ? EC and DE ?
EB.
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