Triangle ABC is isosceles with AB=BC. There is a point D on AC such that BD = DC and AB=AD.

What is the measure of angle BAC?

Right off the bat, since we are told triangle ABC is isosceles, we know that angle BAC = BCA. I will call this angle a. Also, I will call angle ABC, angle b.

Let's look at triangle BDC. Well, since BD = DC, triangle BDC is isosceles. Notice that angle a is in both triangles BDC and ABC (and angle a is not between the two equal legs in either triangle). Therefore, triangles BDC and ABC are similar. That means that angle BDC = angle ABC = angle b.

Also, triangle ABD is isosceles since AB = AD. We've already called angle BAD, angle a. And we'll call angles ABD and ADB, angle c.

Simple relationships:
2a+b = 180 (from triangle ABC or BDC)
2c+a = 180 (from triangle ABD)
c+b = 180 (supplementary angles, since ADC is a straight line = 180)
a+c = b (simple angle addition at angle ABC)

b = 180 – 2a
c = (180 – a)/2

c + b = 180 = (180 – a)/2 + (180 – 2a)
180 = 90 –a/2 +180 – 2a
0 = 90 – 2.5a
2.5a = 90
a = 36

So:
Angle BAC = 36
Angle BCA = 36
Angle ABC = 108
Angle BDE = 108
Angle ABD = 72
Angle ADB = 72

(fixing formatting)

Edited on September 1, 2004, 8:33 am

Posted by nikki on 2004-09-01 08:31:31