Page 134 - Math Course 3 (Book 2)
P. 134
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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