4.1 4A1 BA and 4C1 BC are congruent by SAS. It follows that A1 A ∼
= C1 C.
4.2 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
4BA1 D and 4BC1 D have congruent angles ∠A1 BD and ∠C1 BD and
common hypotenuse. It follows that they are congruent, hence DA1 ∼
=
DB1 .
In the other direction, if DA1 ∼
= DB1 then the right triangles 4BA1 D
and 4BC1 D have a pair of congruent legs and a common hypotenuse,
hence they are congruent. It follows that ∠A1 BD ∼
= ∠C1 BD.
4.3 In 4AA2 B we have AA2 ≤ AB + BA2 . Triangles 4CAA1 and 4BA2 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 ≤ 12 (AB + CA).
←→
←→
4.4 Suppose that on the contrary, the lines AB and CD intersect. Then in
the pair of angles ∠ABC, ∠BCD one will be external and one will be
remote internal of a triangle. But then they can not be equal, and we get
a contradiction.
1