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Difference between revisions of "2020 USOJMO Problems/Problem 4"
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Let <math>FE</math> meet <math>AC</math> at <math>J</math>, <math>BE</math> meet <math>AC</math> at <math>S</math>, connect <math>AE</math>.
 
Let <math>FE</math> meet <math>AC</math> at <math>J</math>, <math>BE</math> meet <math>AC</math> at <math>S</math>, connect <math>AE</math>.
 
Denote that <math>\angle{BCA}=\alpha; AB=BC, \angle{BAC}=\angle{BCA}=\alpha</math>, since <math>EF</math> is parallel to <math>BC</math>, <math>\angle{AJF}=\angle{ACB}=\alpha</math>. <math>\angle{AJF}</math>and <math>\angle{EJS}</math> are vertical angle, so they are equal to each other.  
 
Denote that <math>\angle{BCA}=\alpha; AB=BC, \angle{BAC}=\angle{BCA}=\alpha</math>, since <math>EF</math> is parallel to <math>BC</math>, <math>\angle{AJF}=\angle{ACB}=\alpha</math>. <math>\angle{AJF}</math>and <math>\angle{EJS}</math> are vertical angle, so they are equal to each other.  
−
<math>BE\bot{AC}</math>,\angle{JES}=90^{\circ}-\alpha<math>, since </math>\angle{EFB}=\angle{AJF}+\angle{FAJ}=2\alpha<math>, we can express </math>\angel{FBE}=180^{\circ}-2\alpha-(90^{\circ}-\alpha)=90^{\circ}-\alpha=
+
<math>BE\bot{AC}</math>,<math>\angle{JES}=90^{\circ}-\alpha</math>, since <math>\angle{EFB}=\angle{AJF}+\angle{FAJ}=2\alpha</math>, we can express <math>\angel{FBE}=180^{\circ}-2\alpha-(90^{\circ}-\alpha)=90^{\circ}-\alpha=
−
\angel{FEB}--FE=FB$
+
\angel{FEB}--FE=FB</math>

Revision as of 04:22, 9 January 2022

Problem

Let $ABCD$ be a convex quadrilateral inscribed in a circle and satisfying $DA < AB = BC < CD$. Points $E$ and $F$ are chosen on sides $CD$ and $AB$ such that $BE \perp AC$ and $EF \parallel BC$. Prove that $FB = FD$.

Solution

Let $G$ be the intersection of $AE$ and $(ABCD)$ and $H$ be the intersection of $DF$ and $(ABCD)$.

Claim: $GH || FE || BC$

By Pascal's on $GDCBAH$, we see that the intersection of $GH$ and $BC$, $E$, and $F$ are collinear. Since $FE || BC$, we know that $HG || BC$ as well. $\blacksquare$

Note that since all cyclic trapezoids are isosceles, $HB = GC$. Since $AB = BC$ and $EB \perp AC$, we know that $EA = EC$, from which we have that $DGCA$ is an isosceles trapezoid and $DA = GC$. It follows that $DA = GC = HB$, so $BHAD$ is an isosceles trapezoid, from which $FB = FD$, as desired. $\blacksquare$

Solution 2

Let $G=\overline{FE}\cap\overline{BC}$, and let $G=\overline{AC}\cap\overline{BE}$. Now let $x=\angle ACE$ and $y=\angle BCA$.

From $BA=BC$ and $\overline{BE}\perp \overline{AC}$, we have $AE=EC$ so $\angle EAC =\angle ECA = x$. From cyclic quadrilateral ABCD, $\angle ABD = \angle ACD = x$. Since $BA=BC$, $\angle BCA = \angle BAC = y$.

Now from cyclic quadrilateral ABC and $\overline{FE}\parallel \overline{BC}$ we have $\angle FAC = \angle BAC = \pi - \angle BCD = \pi - \angle FED$. Thus F, A, D, and E are concyclic, and $\angle DFG = \angle DAE = \angle DAC - \angle EAC = \angle DBC - x$ Let this be statement 1.

Now since $\overline{AH}\perp \overline {BH}$, triangle ABC gives us $\angle BAH + \angle ABG = \frac{\pi}{2}$. Thus $y+x+\angle GBE=\frac{\pi}{2}$, or $\angle GBE = \frac{\pi}{2}-x-y$.

Right triangle BHC gives $\angle HBC = \frac{\pi}{2}-y$, and $\overline{BC}\parallel \overline{FE}$ implies $\angle BEG=\angle HBC = \frac{\pi}{2}-y.$

Now triangle BGE gives $\angle BGE = \pi - \angle BEG - \angle GBE = \pi - (\frac{\pi}{2}-y)-(\frac{\pi}{2}-x-y)=x+2y$. But $\angle FGB = \angle BGE$, so $\angle FGB=x+2y$. Using triangle FGD and statement 1 gives \begin{align*}\angle FDG &= \pi - \angle DFG - \angle FGB \\ &= \pi - (\angle DBC - x) - (x + 2y) \\ &= \pi - (\angle GBE + \angle EBC - x) - (x + 2y) \\ &= \pi - ([\frac{\pi}{2}-x-y]+[\frac{\pi}{2}-y]-x)-(x+2y) \\ &= x \\ &= \angle FBD\end{align*}

Thus, $\angle FDB = \angle FBD$, so $\boxed{FB=FD}$ as desired.$\blacksquare$

~MortemEtInteritum

Solution 3 (Angle-Chasing)

Proving that $FB=FD$ is equivalent to proving that $\angle FBD= \angle FDB$. Note that $\angle FBD=\angle ACD$ because quadrilateral $ABCD$ is cyclic. Also note that $\angle BAC=\angle ACB$ because $AB=BC$. $AE=EC$, which follows from the facts that $BE \perp AC$ and $AB=AC$, implies that $\angle CAE= \angle ACE= \angle ACD= \angle FBD$. Thus, we would like to prove that triangle $FBD$ is similar to triangle $AEC$. In order for this to be true, then $\angle BFD$ must equal $\angle AEC$ which implies that $\angle AFD$ must equal $\angle AED$. In order for this to be true, then quadrilateral $AFED$ must be cyclic. Using the fact that $EF \parallel BC$, we get that $\angle AFE= \angle ABC$, and that $\angle FED= \angle BCE$, and thus we have proved that quadrilateral $AFED$ is cyclic. Therefore, triangle $FDB$ is similar to isosceles triangle $AEC$ from AA and thus $FB=FD$.

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Solution 4

BE is perpendicular bisector of AC, so $\angle ACE = \angle EAC$. FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. $\angle AFD = \angle AED = 2 \angle ACE = 2 \angle FBD$. Hence, $\angle FBD = \angle BDF$, $FD = FB$.

Mathdummy

Solution 5

Let $G$ be on $AB$ such that $GD \perp DB$, and $H = GD \cap EB$. Then $\angle{ADB} = \angle{EDB} = 90^{\circ} - \angle{ABE} \implies \triangle{DAE}$ is the orthic triangle of $\triangle{HGB}$. Thus, $F$ is the midpoint of $GB$ and lies on the $\perp$ bisector of $DB$.

Solution 6

Let $FE$ meet $AC$ at $J$, $BE$ meet $AC$ at $S$, connect $AE$. Denote that $\angle{BCA}=\alpha; AB=BC, \angle{BAC}=\angle{BCA}=\alpha$, since $EF$ is parallel to $BC$, $\angle{AJF}=\angle{ACB}=\alpha$. $\angle{AJF}$and $\angle{EJS}$ are vertical angle, so they are equal to each other. $BE\bot{AC}$,$\angle{JES}=90^{\circ}-\alpha$, since $\angle{EFB}=\angle{AJF}+\angle{FAJ}=2\alpha$, we can express $\angel{FBE}=180^{\circ}-2\alpha-(90^{\circ}-\alpha)=90^{\circ}-\alpha= \angel{FEB}--FE=FB$ (Error compiling LaTeX. Unknown error_msg)

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