Clay Kitchings :: EMAT 6600 :: 100 Degree Isosceles Triangle

Given an isosceles triangle ABC with AB = AC and the measure of angle BAC = 100 degrees. Extend AB to point D such that AD = BC. Now draw segment CD. What is the size of angle BCD?

We first know that <ABC and <ACB are congruent, as triangle ABC is isosceles. Furthermore, <ABC and <ACB both measure 40 degrees since < BAC measures 100 degrees.

After constructing segment CD, use it to reflect triangle ABC over itself.  This provides a convenient 90 degree angle where segment AF is perpendicular to CD at point G.  By construction, we now have triangle ADF is equilateral, so each angle in triangle ADF is 60 degrees. We know <FAC is 40 degrees since m<BAC is 100 degrees.  Segments AC and CF are also congruent from the reflection, so we have another convenient isosceles triangle (ACF) with base angles of 40 degrees.  Now we can see that triangle AEB must also be isosceles, since its two base angles (EAB and EBA) measure 40 degrees.

Now, we can see that triangle AGB is a right triangle with two angles measuring 40 degrees and 90 degrees.  Therefore, <ABG must measure 180 – 40 – 90, which is 50 degrees.  That leaves m<CBD = 10 degrees.