Math 367-500
Homework 5
Solutions
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1. Denote by A1 and C1 the ends of the perpendiculars from D to BA and BC, respectively. If D belongs to
the bisector, then the right triangles △BA1 D and △BC1 D have congruent angles ∠A1 BD and ∠C1 BD
and common hypotenuse. By SAA, they are congruent, hence DA1 ∼
= DB1 .
∼
In the other direction, if DA1 = DB1 then the right triangles △BA1 D and △BC1 D have a pair of
congruent legs and a common hypotenuse, hence, by the Hypotenuse-Leg Theorem, they are congruent.
It follows that ∠A1 BD ∼
= ∠C1 BD.
2. In △AA2 B we have AA2 ≤ AB + BA2 . Triangles △CAA1 and △BA2 A1 are congruent by SAS: angles
∠CA1 A and ∠BA1 A2 form a vertical pair, CA1 ∼
= BA1 , and AA1 ∼
= A1 A2 . Consequently, BA2 ∼
= CA,
and we get 2AA1 ≤ AB + CA, hence AA1 ≤ 21 (AB + CA).
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3. BC is a transversal for AB and CD. Then ∠ABC and ∠BCD are alternate interior angles. Since
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∠ABC ∼
= ∠BCD, then, by Theorem 3 (section 10.1), AB∥CD.
4. Let ∠X be an angle between the lines T and L1 , and ∠Y be an angle between the lines T and L2 . Since
T is perpendicular L1 , then m∠X = 90. Since L1 and L2 are parallel, then, by Theorem 1 (section 11.1),
∠X ∼
= ∠Y , therefore, m∠Y = 90. Thus, L2 is perpendicular to T .
5. Given that L1 ∥L and L2 ∥L. Prove that L1 ∥L2 . We prove by contradiction. Suppose, L1 ∥L and L2 ∥L,
but L1 and L2 are not parallel. Then L1 and L2 intersect at a point P . We have two lines that pass
through the point P and are parallel to L, which contradicts P-1.
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