Constructing triangles holding many similarities

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Constructing triangles holding many similarities

G H J

Reveal topic tags

geometry proposed

Isogonal conjugate

L

G H BBookmark kLocked kLocked NReply

Source: Baltic Way 2011

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1347 posts

WakeUp

#1 h Nov 6, 2011, 1:10 PM • 2 Y

Y by Adventure10, Mango247

Let $E$ be an interior point of the convex quadrilateral $ABCD$ . Construct triangles $\triangle ABF,\triangle BCG,\triangle CDH$ and $\triangle DAI$ on the outside of the quadrilateral such that the similarities $\triangle ABF\sim\triangle DCE,\triangle BCG\sim \triangle ADE,\triangle CDH\sim\triangle BAE$ and $\triangle DAI\sim\triangle CBE$ hold. Let $P,Q,R$ and $S$ be the projections of $E$ on the lines $AB,BC,CD$ and $DA$ , respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then
$EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

418 posts

skytin

#2 h Nov 6, 2011, 1:40 PM • 2 Y

Y by Adventure10, Mango247

Hints :
Let E' is isogonal to E wrt ABC
E'BCD ~ AFEI , AFBE , IAED , EBGC , DECH are cycic

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

134 posts

KRIS17

#3 h Aug 18, 2019, 10:01 PM • 1 Y

Y by Adventure10

Based on the given similarities of the triangles, it is easy to see that quadrilaterals $AFBE$ and $DECH$ are similar.
Thus, we have:
$AB/DC$ = $FE/EH$
or $AB.EH$ = $FE.DC$

Same way, since quadrilaterals $DIAE$ and $CEBG$ are similar, we get:
$EG.DA$ = $EI.BC$

It remains to be proven that:
$EG.DA$ = $AB.EH$ as the fourth eqiality follows directly form these 3 eualities.

Since quadrilateral $PQRS$ is cyclic, we get that:
$\angle RSE$ = $\angle RDE$ = $\angle BAF$ (from similarities given)
and $\angle ESP$ = $\angle EAP$
Thus, we have $\angle RSP$ = $\angle EAF$ .

In the same manner we can derive that $\angle PQR$ = $\angle FBE$ .

Since, $\angle RSP$ and $\angle PQR$ are supplementry, so are $\angle EAF$ and $\angle FBE$ .
Thus, quadrilateral $AFBE$ is cyclic. Since quadrilaterals $AFBE$ and $DECH$ are similar, $DECH$ is cyclic too.
In the same manner, we can derive that, quadrilaterals $DIAE$ and $CEBG$ are cyclic too.

Applying Ptolemy's theorem to $AFBE$ , we get:
$AB.EF$ = $AF.EB$ + $FB.AE$
or $AB$ = $(AF/EF).EB + (FB/EF).AE$ = $(DE/EH).EB + (EC/EH).AE$ (from similarity of quadrilaterals $AFBE$ and $DECH$ )
or $AB.EH$ = $DE.EB$ + $EC.AE ---> (1)$

In a similar manner, by applying Ptolemy's theorem to $DIAE$ , we get:
$EG.AD$ = $DE.EB$ + $EC.AE ---> (2)$

From $(1)$ and $(2)$ , we have $EG.DA$ = $AB.EH$ which completes the proof.

This post has been edited 1 time. Last edited by KRIS17, Aug 18, 2019, 10:03 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1079 posts

#4 h Aug 13, 2021, 9:10 PM

Y by

Claim. $ABEF$ , $BECG$ , $CEDH$ and $AEDI$ are cyclic.
Proof. We have
$\begin{align*}\measuredangle IAE&=\measuredangle SAE+\measuredangle EBQ\\&=\measuredangle SPE+\measuredangle EPQ\\&=\measuredangle SPQ\\&=\measuredangle SRQ\\&=\measuredangle SRE+\measuredangle ERQ\\&=\measuredangle SDE+\measuredangle ECQ\\&=\measuredangle SDE+\measuredangle IDA\\&=\measuredangle IDE, \end{align*}$ we do the same for others, the claim follows.

Claim.
There exists $T$ , the isogonal conjugate of $E$ wrt $ABCD$ .
Proof. By six-point circle theorem, we conclude that there exists $T$ , the isogonal conjugate of $E$ .

Now, note that angle chasing implies that $\triangle GBE\sim\triangle CTD$ and $\triangle FBE\sim\triangle ATD$ , thus
$\frac{EF}{EG}=\frac{AD}{DC}.$ Similarly, $\dfrac{EG}{EH}=\dfrac{AB}{AD}$ , $\dfrac{EH}{EI}=\dfrac{BC}{AB}$ and $\dfrac{EI}{EF}=\dfrac{CD}{BC}$ .
This yields that
$EF\cdot CD=EG\cdot DA=EH\cdot AB=EI\cdot BC.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

117 posts

#5 h Apr 15, 2025, 1:37 PM

Y by

Here is a solution using spiral similarity and avoiding isogonal conjugates.

$EF \cdot CD= EH \cdot AB$ is clear since there is similar quadrilaterals. So let's prove that $EF \cdot CD=EG \cdot DA$ .

We have $\angle FEG=\angle FEB+\angle BEG=\angle FAB+\angle BCG= \angle EDC+\angle EDA=\angle ADC$ . Now let's consider $S$ , the center of the spiral similarity that takes $AB$ to $CD$ . So it takes $F$ to $E$ . We have $\frac{EF}{ES}=\frac{AD}{DS}$ .
Since $S$ is also the center of the similarity that takes $AD$ to $BC$ , we have $\frac{EG}{ES}=\frac{CD}{DS}$ .
Combining these two relations we get $\frac{AD}{CD}=\frac{EF}{EG}$ , So $\triangle FEG \sim \triangle ADC$ . Conclusion follows.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a