Let $ABC$ be a triangle with $\angle C=60^\circ$ and $AC<BC$. The point $D$ lies on the side $BC$ and satisfies $BD=AC$. The side $AC$ is extended to the point $E$ where $AC=CE$. Prove that $AB=DE$.
Problem
Source: Baltic Way 1999
Tags: trigonometry, geometry, geometric transformation, rotation, geometry proposed
uglysolutions
01.09.2008 22:55
Trivial by cosine theorem. Let $ AC = CE = BD = y$ and $ CD = x$. In triangle $ ECD$: $ ED^2 = x^2 + y^2 - 2xy \cos 120 = x^2 + y^2 + xy$. In triangle $ ACB$: $ AB^2 = x^2 + (x + y)^2 - 2x(x + y) \cos 60 = x^2 + x^2 + 2xy + y^2 - x(x + y) = x^2 + x^2 + 2xy + y^2 - x^2 - xy = x^2 + y^2 + xy$. Hence $ AB = DE$ as wanted.
mr.danh
02.09.2008 05:01
anonymous111 wrote:
Let $ ABC$ be a triangle with $ \angle C = 60^\circ$ and $ AC < BC$.The point $ D$ lies on the side $ BC$ and satisfies $ BD = AC$.The side $ AC$ is extended to the point $ E$ where $ AC = CE$.Prove that $ AB = DE$
Let K on CB such that CK=CA, so triangle ACK is equilateral. Since CK=CA=BD, BK=CD, $ \angle AKB=\angle ECD=120^o$, we have that $ \triangle AKB=\triangle ECD$, hence DE=AB.
sunken rock
27.10.2008 08:32
Draw the circle (ABC), name P the middle of the arc AB containing C. Since a 60 degs clockwise rotation about P maps the triangles PCA and PDB, hence PC = PD = CD and <PDB = 120 degs, so the triangles PDB and DCE are equal, hence DE = PB = AB. Best regards, sunken rock