ABC is an isosceles triangle with AB=AC. Point D lies on side BC. Point F is inside $\triangle$ABC and lies on the circumcircle of triangle ADC. The circumcircle of triangle BDF intersects side AB at point E. Prove that $CD\cdot EF+DF\cdot AE=BD\cdot AF$.
Problem
Source: China south east mathematical olympiad 2004 day2 problem 6
Tags: geometry, circumcircle, trigonometry, cyclic quadrilateral, geometry unsolved
exmath89
30.06.2013 05:06
We invert the diagram through a circle, $\omega$, centered at $A$ with radius $r$ such that $\omega$ is orthogonal to $(BDF)$. Denote $K'$ as the inverse of an object $K$. Note that the inverse of $(BDF)$ is itself, so $B'=E$ and $E'=B$. Also, $D'$ and $F'$ lie on $(BDF)$. Since $ACDF$ is cyclic, $C',D',F'$ are collinear. Because $B,C,D$ are collinear, $AB'D'C'$ is cyclic. Using the Inversion Distance Formula, it suffices to prove
\[ \frac{r^2*C'D'}{AC'*AD'} *\frac{r^2*E'F'}{AE'*AF'}+\frac{r^2*D'F'}{AD'*AF'}*\frac{r^2}{AE'})=\frac{r^2*B'D'}{AB'*AD'}*\frac{r^2}{AF'}\]
which, after observing that $AB=AC\implies AB'=AC'$, can be simplified to
\[ C'D'*E'F'+D'F'*AC'=B'D'*AE'.\]
Next, note that by construction, $DD'E'F'$ and $CC'D'D$ are cyclic. Thus, $\angle{DE'F'}=\angle{DD'F}=\angle{C'CD}$, implying that $E'F'||AC$. Let lines $AE'$ and $C'F'$ intersect at $X$. Then $\triangle{XE'F'}\sim\triangle{XAC'}$ by AA similarity, so
\[ \frac{XF'}{XD'+C'D'}=\frac{E'F'}{AC'}\implies XF'*AC'=XD'*E'F'+C'D'*E'F'\]
Since $AC'D'B'$ is cyclic, $\angle{AC'X}=\angle{D'B'X}$ and $\angle{C'AX}=\angle{B'D'X}$, so $\triangle{XB'D'}\sim\triangle{XC'A}$ by AA similarity. Therefore,
\[ \frac{XF'+D'F'}{AE'+XE'}=\frac{B'D'}{AC'}\implies XF'*AC'+D'F'*AC'=B'D'*AE'+B'D'*XE'\]
Finally, from cyclic quadrilateral $B'D'F'E'$, we have $\angle{XE'F'}=\angle{XD'B'}$ and $\angle{XF'E'}=\angle{XB'D'}$. Hence, $\triangle{XE'F'}\sim\angle{XD'B'}$ by AA similarity, and
\[ \frac{XE'}{XD'}=\frac{E'F'}{B'D'}\implies B'D'*XE'=XD'*E'F'\]
As a result, substituting the first equation into the second and then the third into the second, we find that
\[ C'D'*E'F'+D'F'*AC'=B'D'*AE'\]
as desired. //
elegant
06.07.2013 01:54
Please give me a solution without using Inversion.
Arab
06.07.2013 04:33
Denote by $K$ the intersection of $AF$ and the circumcircle of $\triangle BDF$.Obviously,$\triangle AEF\sim \triangle ABK$,so $\frac{EF}{AF}=\frac{BK}{AB},\frac{AE}{AF}=\frac{AK}{AB}$. hence $CD\cdot EF+DF\cdot AE=BD\cdot AF \iff CD\cdot BK+DF\cdot AK=BD\cdot AB$. Note that,$\angle KBD=\angle KFD=\angle ACB$,we have $[DCK]=\frac{1}{2}\cdot CD\cdot BK\cdot \sin{\angle KBD}=\frac{1}{2}\cdot CD\cdot BK\cdot \sin{\angle ACB}$, $[ADK]=\frac{1}{2}\cdot AK\cdot DF\cdot \sin{\angle ACB}$, $[ABD]=\frac{1}{2}\cdot BD\cdot AB\cdot \sin{\angle ACB}$. Since $\angle KBD=\angle ACB$ implies that $B\parallel AC$,we have $[ABC]=[AKC],[ABD]=[DCK]+[ADK]$. Consequently,$CD\cdot EF+DF\cdot AE=BD\cdot AF$. $Q.E.D.$
N0_NAME
31.07.2020 22:14
Denote by $X$ the intersection of $AB$ and the circumcircle of $\triangle ADC$. Since $\angle{CXF}=\angle{BDF}=\angle{AEF}$ and $\angle{XCF}=\angle{EAF}$ implies that $\triangle{AEF}\sim\triangle{CFX}$. We have $k=\frac{AE}{CX}=\frac{AF}{CF}=\frac{EF}{XF}$. $\angle{ABC}=\angle{DCA}=\angle{DXB}\implies BD=DX$. Ptolemy's theorem:$CD\cdot XF+DF\cdot CX=DX\cdot CF \implies CD\cdot XF\cdot k+DF\cdot CX\cdot k=BD\cdot CF\cdot k \implies$ $CD\cdot EF+DF\cdot AE=BD\cdot AF$.