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In convex $\square ABCD$, $\angle BAD=\angle CDA$

The midpoints of $AB$,$CD$,$DA$ are $L$,$M$,$N$ respectively.

$\overline{AC}$ meet $\overline{BD}$ at point $E$.

Let $w$ be a circle that passes through $E$ and is tangent to $\overset{\longleftrightarrow}{AD}$ at point $A$.

Let $\overline{NE}$ cut $w$ at point $F$, which is not $E$.

Show that $\angle LFE=\angle MFE$

enter image description here

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  • $\begingroup$ my trials are angle bisector theorem, harmonic conjugate, congruent triangles etc...but I couldn't find any clues in my trials. $\endgroup$ – chloe_shi Nov 13 '14 at 3:49
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Consider the most extreme case: if $ABCD$ were to be square.

enter image description here

If that were the case, $L$ and $M$ would lie on the same line, the midpoint of the chord $AE$ is $J$, and with $AD\perp AB$, this happens to show that $M=w$. If $ABCD$ is a square, then $E, L$ and $M$ will be co-linear. Since $AM=AE$, it goes to show that $F=E$, and by pure consequence $\angle LEF$ and $\angle MEF$ are both $180^\circ$, thus equal.


Suppose $ABCD$ is a rectangle.

W.l.o.g., although this holds true too for all rectangles, assume that $AD=BC=s$ and $CD=AB=2s$. Inevitably, the center of circle $w$ will lie on $AB$ since $AB\perp AD$

If you take some time to calculate, you'll get the following lengths and angles: $$\begin{align} Aw&=\frac{3\sqrt{20}}{20}s\\ FA&=\frac{\sqrt{30}}{10}s\\ NF&=\frac{\sqrt5}{10}s\\ FE&=s-NF=\frac{10-\sqrt5}{10}s\\ \end{align}$$

For angles $\angle LFE$ and $\angle MFE$, since $L, E$ and $M$ are co-linear and thus create right triangles with $F$, you get $$\tan LFE=\frac{EL}{EF}\Rightarrow LFE=\tan^{-1}\frac{EL}{EF}$$ $$\tan MFE=\frac{EM}{EF}\Rightarrow MFE=\tan^{-1}\frac{EM}{EF}$$

Since they both angles share $FE$ and $EL=EM=\frac12s$, we are able to prove that indeed: $$\angle LFE=\angle MFE$$

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