Consider triangles ABC, A B C such that AB = A B , AC = A C , and
A A (Fig. 30). We want to prove that BC B C . Let us move the second
triangle so that A B coincides with AB. Since A is less than A, side A C will
move to a position AD in the interior of angle BAC. We construct the bisector
AE of DAC. This segment is also inside BAC, and, as B and C lie on different
sides of this line, it must intersect side BC in some point E located between B and
C. If we draw DE, we see that triangles ACE and ADE are congruent, because
they have an equal angle (with vertex A) between two corresponding equal sides
(AE in common, AC = A C = AD). Therefore DE = EC. The inequality
BD BE + ED provided by the triangle BDE then gives us
Conversely. If, in two triangles, two pairs of sides are equal, but the third
sides are unequal, then the angles opposite the unequal sides are unequal, and the
greater angle is opposite the greater side.
This statement is equivalent to the preceding one.
Remark. The preceding theorem does not require the third case of congruence
for triangles (SSS). It therefore provides another proof of that case.
Indeed, if we have AB = A B , AC = A C and, in addition BC = B C ,
the angles A and A would have to be equal, or else BC and B C could not be
equal. Knowing now that A = A the two triangles are congruent by the second
case (SAS).
Exercise 5. Prove that a triangle is isosceles:
1◦. if an angle bisector is also an altitude;
2◦. if a median is also an altitude;
3◦. if an angle bisector is also a median.
Exercise 6. On side Ox of some angle, we take two lengths OA, OB, and on
side Ox we take two lengths OA , OB , equal respectively to the first two lengths.
We draw AB , BA , which cross each other. Show that point I, where these two
segments intersect, lies on the bisector of the given angle.
Exercise 7. If two sides of a triangle are unequal, then the median between
these two sides makes the greater angle with the smaller side. (Imitate the con-
struction in 25.)
Exercise 8. If a point in the plane of a triangle is joined to the three vertices
of a triangle, then the sum of these segments is greater than the semi-perimeter
of the triangle; if the point is inside the triangle, the sum is less than the whole
Exercise 8b. If a point in the plane of a polygon is joined to the vertices of
the polygon, then the sum of these segments is greater than the semi-perimeter of
the polygon.
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