# Properties of an Isosceles Triangle

Definition: A triangle is isosceles if two of its sides are equal.

We want to prove the following properties of isosceles triangles.

 Theorem: Let ABC be an isosceles triangle with AB = AC.  Let M denote the midpoint of BC (i.e., M is the point on BC for which MB = MC).  Then a)      Triangle ABM is congruent to triangle ACM. b)      Angle ABC = Angle ACB (base angles are equal) c)      Angle AMB = Angle AMC = right angle. d)      Angle BAM = angle CAM

Corollary: Consequently, from these facts and the definitions:

• Ray AM is the angle bisector of angle BAC.
• Line AM is the altitude of triangle ABC through A.
• Line AM is the perpendicular bisector of B
• Segment AM is the median of triangle ABC through A.

### Proof #1 of Theorem (after B&B)

Let the angle bisector of BAC intersect segment BC at point D.

• Also, AB = AC since the triangle is isosceles.

This means that triangle BAD = triangle CAD, and corresponding sides and angles are equal, namely:

1. DB = DC,
2. angle ABD = angle ACD,

(Proof of a).  Since DB = DC, this means D = M by definition of the midpoint.  Thus triangle ABM = triangle ACM.

(Proof of b) Since angle ABD = angle ABC (same angle) and also angle ACD = angle ACB, this implies angle ABC = angle ACB.

(Proof of c) From congruence of triangles, angle AMB = angle AMC.  But by addition of angles, angle AMB + angle AMC = straight angle = 180 degrees.  Thus 2 angle AMB = straight angle and angle AMB = 90 degrees = right angle.

(Proof of d) Since D = M, the congruence angle BAM = angle CAM follows from the definition of D.  (These are also corresponding angles in congruent triangles ABM and ACM.)

QED*

*Note:  There is one point of this proof that needs a more careful “protractor axiom”.  When we constructed the angle bisector of BAC, we assumed that this ray intersects segment BC.  This can’t be quite deduced from the B&B form of the axioms.  One of the axioms needs a little strengthening.

The other statements are immediate consequence of these relations and the definitions of angle bisector, altitude, perpendicular bisector, and median.  (Look them up!)

Definition:  We will call the special line AM the line of symmetry of the isosceles triangle.  Thus we can construct AM as the line through A and the midpoint, or the angle bisector, or altitude or perpendicular bisector of BC. Shortly we will give a general definition of line of symmetry that applies to many kinds of figure.

### Proof #2 (This is a slick use of SAS, not presented Monday.  We may discuss in class Wednesday.)

The hypothesis of the theorem is that AB = AC.  Also, AC = AB (!) and angle BAC = angle CAB (same angle).  Thus triangle BAC is congruent to triangle BAC by SAS.

The corresponding angles and sides are equal, so the base angle ABC = angle ACB.

Let M be the midpoint of BC.  By definition of midpoint, MB = MC. Also the equality of base angles gives angle ABM = angle ABC = angle ACB = angle ACM.  Since we already are given BA = CA, this means that triangle ABM = triangle ACM by SAS.

From these congruent triangles then we conclude as before:

• Angle BAM = angle CAM (so ray AM is the bisector of angle BAC)
• Angle AMB = angle AMC = right angle (so line MA is the perpendicular bisector of  BC and also the altitude of ABC through A)

QED

### Faulty Proof #3.  Can you find the hole in this proof?)

In triangle ABC, AB = AC.  Let M be the midpoint and MA be the perpendicular bisector of BC.

• Then angle BMA = angle CMA = right angle, since MA is perpendicular bisector.
• MB = MC by definition of midpoint. (M is midpoint since MA is perpendicular bisector.)
• AM = AM (self).

So triangle AMB = triangle AMC by SAS.

Then the other equal angles ABC = ACB and angle BAM = angle CAM follow from corresponding parts of congruent triangles.  And the rest is as before.

QED??