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?
(In reply to Solution + Explanation
<<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.>>
Actually, where did AB = AD come from? By symmetry, since BD = DC, it is BD = AD. And since angle ABD equals angle a, and angle CBD = angle a, and they add up to angle ABC, b=2a. But b+2a = 180 degrees. So b = 90 degrees and a = 45 degrees.
Posted by Charlie
on 2004-09-01 08:43:45