Page 134 - Math Course 3 (Book 2)

Proportional Parts of Triangles

Triangle ABC has vertices A(–2, 2), B(2, 4) and First, use the Distance Formula to find BC and DE.
C(4, –4). DE is a midsegment of △ABC.
Verify that BC || DE. BC = (2 – 4)² + [4 –( –4)]²

B = 4 + 64
y
D (2, 4)
= 68 or 2 17

(–2, 2)
A
DE = (0 – 1)² + [3 –( –1)]²

= 1 + 16
O x

E = 17

DE √17 1
(4, –4) BC = 2 √17 or 2
C

1
1
DE
If the slopes of BC and DE are equal, BC || DE. Answer If = , then DE = BC.
2
2
BC
– 4 – 4
slope of BC = or – 4
4 – 2
–1 – 3
slope of DE = or – 4
1 – 0
Congruent Segments
Because the slopes BC and DE
Answer
of BC || DE.
Example
Find x and y.
Triangle ABC has vertices A(–2, 2), B(2, 4) and
C(4, –4). DE is a midsegment of △ABC.
1
Verify that DE = BC.
2 2x + 2 3x – 4
B
y
D (2, 4)

(–2, 2)
A 8 y + 7
3
5y
O x
E
To find x:

(4, –4) 2x + 2 = 3x – 4 Given
C 2 = x – 4 Subtract 2x from each side.

6 = x Add 4 to each side.

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