EXERCISES 29

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

BD BE + EC,

or

BD BC.

QED

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).

Exercises

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

perimeter.

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.