Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\omega$. Point $D$ lies on arc $\frown{BC}$ not containing $A$. Let $E$ be the foot of perpendicular from $A$ to line $CD$. Prove that $BC+DC=2DE$.
Problem
Source: MOP 2005 Homework - Red Group #15
Tags: geometry unsolved, geometry
thugzmath10
06.05.2014 18:37
$BD+CD=2DE$.
Let $M$ be the midpoint of $BC$. Then we have $\angle{ABM}=\angle{ABC}=\angle{ADC}=\angle{ADE}$, so $\triangle AMB\sim\triangle AED$, which gives $\frac{MB}{DE}=\frac{AB}{AD}\implies\frac{BC}{2DE}=\frac{AB}{AD}$.
By Ptolemy's Theorem we have $AB(CD+BD)=BC\cdot AD$ (since $AB=AC$), so we get $2DE=\frac{BC\cdot AD}{AB}=CD+BD$.
jayme
07.05.2014 11:52
Dear Mathlinkers, this problem appears as opposite version of the Archimedes broken chord with the antipole of A... The result is then immediat... Sincerely Jean-Louis
BBAI
07.05.2014 12:32
Assuming to prove $BD+DC = 2DE$ Solution : Let $F$ be the point of reflecton of $D$ wrt $E$ on the line $DC$ So $AD=AF$ as $AE \perp CD$. Now in $ \triangle ABD $ and $ \triangle ACF $ , $AD = AF$ $ \angle ACF = \angle ABD $( cyclic) and $AB=AC$ , so both are congruent , so $CF = BD$ Therefore $2DE = DF = CD + CF = CD + BD $ , hence proved
Konigsberg
07.05.2014 19:08
no, it is not $BD+CD=2DE$. It is the original statement (or else there is a mistake on the MOP Homework PDF)
math_science
10.06.2014 08:37
Konigsberg wrote: Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\omega$. Point $D$ lies on arc $\frown{BC}$ not containing $A$. Let $E$ be the foot of perpendicular from $A$ to line $CD$. Prove that $BC+DC=2DE$. Isn't it same as this problem http://www.artofproblemsolving.com/Forum/viewtopic.php?p=358010&sid=b34d670aa2770b67dc7a1f1d3c5dca7d#p358010