if G is the midpoint of FH, then E is the midpoint of GH, cyclic ABCD related

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if G is the midpoint of FH, then E is the midpoint of GH, cyclic ABCD related

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Source: Finland 2019 p3

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#1 h Sep 8, 2019, 3:03 AM • 2 Y

Y by Adventure10, ItsBesi

Let $ABCD$ be a cyclic quadrilateral whose side $AB$ is at the same time the diameter of the circle. The lines $AC$ and $BD$ intersect at point $E$ and the extensions of lines $AD$ and $BC$ intersect at point $F$ . Segment $EF$ intersects the circle at $G$ and the extension of segment $EF$ intersects $AB$ at $H$ . Show that if $G$ is the midpoint of $FH$ , then $E$ is the midpoint of $GH$ .

This post has been edited 1 time. Last edited by parmenides51, May 30, 2020, 9:57 PM
Reason: typo: segment EF instead of line EF

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#2 h Sep 8, 2019, 4:50 AM • 3 Y

Y by AFSA, Adventure10, Mango247

just see the power of point $H$ in $\odot$ $ABF$ and $\odot$ $ABG$

This post has been edited 3 times. Last edited by Vaddokx, Sep 8, 2019, 4:53 AM

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#3 h Jan 20, 2022, 8:08 PM

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Let $\omega$ be circle of $ABCD$ . Draw tangent line to $\omega$ at $A$ and take any point $T$ on this line. Since $GF=GH$ we get $A(T,G;F,H)=-1$ . Projecting it to $\omega \implies (A,G;D,B)=-1$ . Similarly $(B,G;A,C)=-1$ and projecting it to line $FH$ through $A$ gives $EG=EH$ as desired.

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#4 h Dec 11, 2022, 12:12 PM

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Complex numbers...

Let $(ABCD)$ be unit circle and $a=1 , b=-1$
$e=\frac{d+2cd-c}{c+d} \wedge \bar{e}=\frac{c+2-d}{c+d}$ $f=\frac{c+2cd-d}{c+d} \wedge \bar{f}=\frac{d+2-c}{c+d}$ Since $H$ is intersection of $AB$ and $FE$
$h=\frac{\bar{e}f-e\bar{f}}{(f-e)-(\bar{f}-\bar{e})}=\frac{cd+1}{c+d}$ Given that $G$ is midpoint of $FH$ , Hence
$g=\frac{1}{2}(\frac{c+2cd-d}{c+d}+\frac{cd+1}{c+d})=\frac{c+3cd+1-d}{2(c+d)} \wedge \bar{g}=\frac{d+3+cd-c}{2(c+d)}$ Using of the fact that $G$ lies on $(ABCD)$
$ABGD \text{ is cyclic} \iff \frac{g-d}{(g-1)(d+1)}=\frac{\bar{g}-\bar{d}}{(\bar{g}-1)(\bar{d}+1)} \iff (g-d)(\bar{g}-1)-d(\bar{g}-\bar{d})(g-1)=0 \iff -3cd^2-5d^2+8cd+8d-5c-3=0(\clubsuit)$ $ABCG \text{ is cyclic}\iff \frac{c-g}{(c-1)(g+1)}=\frac{\bar{c}-\bar{g}}{(\bar{c}-1)(\bar{g}+1)} \iff (g-c)(\bar{g}+1)-c(\bar{c}-\bar{g})(g+1)=0 \iff -3c^2d+5c^2-8cd+8c-5d+3=0(\diamondsuit)$
$(\clubsuit)$ and $(\diamondsuit)$ yields:
$(c+d)(5c-5d-3cd+3)=0 \text{ if c+d=0 extensions of AD and BC will not intersect , Hence } 5c-5d-3cd+3=0(\spadesuit)$
Now Let us finish problem
$E \text{ is midpoint of } GH \iff g+h=2e\iff \frac{c+3cd+1-d}{2(c+d)}=\frac{2(d+2cd-c)}{c+d} \iff c+5cd-d+3=4d+8cd-4c \iff 5c-5d-3cd+3=0$ And it is correct by $(\spadesuit)$ .....so we are done :-D

:-D

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#5 h Oct 6, 2024, 10:36 AM

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Just use Ratio Lemma and just Trig Bash the problem

This post has been edited 1 time. Last edited by ItsBesi, Oct 6, 2024, 1:59 PM

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