Let $ ABC$ be an isosceles triangle with $ AB=AC$ and $ \angle A=20^\circ$. On the side $ AC$ consider point $ D$ such that $ AD=BC$. Find $ \angle BDC$.
Problem
Source: JBMO Shortlist 2002
Tags: geometry, geometry proposed
jgnr
15.11.2008 16:13
This is a very well-known problem.
Take point $ E$ on the opposite of $ D$ with respect to $ AB$ such that $ \triangle EDA\cong\triangle ABC$.
Note that $ \angle BAE=60^{\circ}$ and hence $ \triangle ABE$ is equilateral. Also note that $ \angle DEB=40^{\circ}$. Since $ ED=EB$, we have $ \angle EDB=70^{\circ}$.
It is easy to see that $ \angle BDE=30^{\circ}$.
sunken rock
23.04.2009 14:24
Take F inside the triangle so that BCF is equilateral, F belongs to the bisector of the angle <BAC, hence <BAF = 10 degs ( 1 ) and <ABF = <ABC - <FBC = 20 degs ( 2 ), then see that triangles ABF and BAD are congruent (s.a.s.), thence <ABD = <BAF = 10 degs and <BDC = 30 degs, q.e.d. Best regards, sunken rock
