Iran(Second Round) 2015,second day,problem 4

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Source: Iran(Second Round) 2015,second day,problem 4

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#1 h May 8, 2015, 10:59 AM • 4 Y

Y by Miku_, lazizbek42, Adventure10, Mango247

In quadrilateral $ABCD$ , $AC$ is bisector of $\hat{A}$ and $\widehat{ADC}=\widehat{ACB}$ . $X$ and $Y$ are feet of perpendicular from $A$ to $BC$ and $CD$ ,respectively.Prove that orthocenter of triangle $AXY$ is on $BD$ .

This post has been edited 1 time. Last edited by MRF2017, May 8, 2015, 11:37 AM

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#3 h May 8, 2015, 11:16 AM • 8 Y

Y by Uagu, reveryu, SAT1001, enhanced, hakN, lazizbek42, Adventure10, Mango247

MRF2017 wrote:

In quadrilateral $ABCD$ , $AC$ is bisector of $\hat{A}$ and $\widehat{ADC}=\widehat{ACB}$ . $X$ and $Y$ are feet of perpendicular from $A$ to $BC$ and $CD$ ,respectively.Prove that orthocenter of triangle $AXY$ is on $B \color{red}{ D }\normalcolor$ .

Typo corrected

My solution :

Let $Z$ be the point on $BD$ such that $XZ \parallel CD$ .

Easy to see $C$ is the antipode of $A$ in $\odot (AXY)$ . ... $(\star)$

From $\angle BAC=\angle CAD, \angle ACB=\angle ADC \Longrightarrow \triangle ABC \sim \triangle ACD$ ,
so $BZ:ZD=BX:XC=CY:YD \Longrightarrow YZ \parallel BC \Longrightarrow XZYC$ is a parallelogram .

Since $Z$ is the reflection of $C$ in the midpoint of $XY$ ,
so combine $(\star)$ we get $Z$ is the orthocenter of $\triangle AXY$ .

Q.E.D

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#4 h May 8, 2015, 1:39 PM • 7 Y

Y by Uagu, phantranhuongth, aritrads, Miku_, hakN, Adventure10, Mango247

MRF2017 wrote:

In quadrilateral $ABCD$ , $AC$ is bisector of $\hat{A}$ and $\widehat{ADC}=\widehat{ACB}$ . $X$ and $Y$ are feet of perpendicular from $A$ to $BC$ and $CD$ ,respectively.Prove that orthocenter of triangle $AXY$ is on $BD$ .

My solution :
//cdn.artofproblemsolving.com/images/b9383f986f3399d20375156e31266ab3e5867b00.jpg

//cdn.artofproblemsolving.com/images/b9383f986f3399d20375156e31266ab3e5867b00.jpg

Easy to see that the circle $(AXY)$ through $C$ .
Let $E,F$ be the intersection of $(AXY)$ and $AB,AC$ respectively.
Let $H$ be the intersection of $EY$ and $FX$ . We apply Pascal theorem for six points $H,A,G,X,C,Y$ , we get that $B,H,D$ are conlinear.
On the other hand,
$\angle AFX=\angle ACX=\angle ADC$ .
So we have $XF$ is parrallel to $C$ , implies $XF \perp AY$ . Similary, $YE \perp AX$ .
Consequenlty, $H$ be the orthocenter of triangle $AXY$ .
We are done.

This post has been edited 3 times. Last edited by Jul, May 8, 2015, 2:02 PM
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hoangtulanhlung19555

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hoangtulanhlung19555

#6 h Jul 4, 2015, 6:24 AM • 3 Y

Y by Vanescralet, Adventure10, Mango247

MRF2017 wrote:

In quadrilateral $ABCD$ , $AC$ is bisector of $\hat{A}$ and $\widehat{ADC}=\widehat{ACB}$ . $X$ and $Y$ are feet of perpendicular from $A$ to $BC$ and $CD$ ,respectively.Prove that orthocenter of triangle $AXY$ is on $BD$ .

My solution:

Let $H$ is orthocenter of triangle $AXY$ .
Easy to see that $XHYC$ is parallelogram.
Thus,
$\widehat{BXH}=\widehat{HYD}$
Now we have to prove that $\dfrac{BX}{HX}=\dfrac{HY}{YD}$ .

It's equivalent to $\dfrac{BX}{CY}=\dfrac{CX}{YD}$ .
Easy to see that triangle $BAC$ is similar to triangle $CAD$ .
So that triangle $BAX$ is simliar to triangle $CAY$ , triangle $CAX$ is similar to triangle $DAY$ , thus $\dfrac{BX}{CY}=\dfrac{AX}{AY}$ and $\dfrac{CX}{YD}=\dfrac{AX}{AY}$ , therefore $\dfrac{BX}{CY}=\dfrac{CX}{YD}$ .
Thus, triangle $BXH$ is similar to triangle $HYD$ . So,
$\widehat{BHX}+\widehat{XHY}+\widehat{YHD}=\widehat{BHX}+\widehat{BXH}+\widehat{XBH}=180^o$
Hence, $B, H, D$ are collinear.
We are done.

Attachments:

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#7 h Jul 4, 2015, 8:58 AM • 2 Y

Y by Adventure10, Mango247

Let $H$ be the intersection point of $BD$ with the perpendicular from $X$ to $AY$ . Then $HX \parallel DC$ , so $\frac{BH}{HD}=\frac{BX}{XC} \displaystyle {(1)}$ .

Triangles $ADY$ and $ACX$ are similar because they are right and $\angle ADY= \angle ACX$ .
Hence $\frac{DY}{XC}=\frac{AY}{AX} \displaystyle {(2)}$ .

Also triangles $AYC$ and $AXB$ are similar because they are right and $\angle YAC= \angle DAC - \angle DAY = \angle CAB - \angle CAX= \angle BAX$ .
Therefore $\frac{YC}{BX}=\frac{AY}{AX} \displaystyle {(3)}$ .

By $\displaystyle {(2)}$ and $\displaystyle {(3)}$ it is $\frac{DX}{XC}=\frac{YC}{BX} \Rightarrow \frac{BX}{XC}=\frac{CY}{YD}$ .

Then by $\displaystyle {(1)}$ it is $\frac{DH}{HB}=\frac{DY}{YC}$ . Thus, we obtain that $HY \parallel BC$ .

But $AX\perp BC$ yields that $HY\perp AX$ .
So $H$ is the orthocenter of triangle $AXY$ and it lies on $BD$ .

This post has been edited 1 time. Last edited by eudoxos99, Jul 4, 2015, 9:03 AM

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#8 h Sep 29, 2019, 7:50 PM • 1 Y

Y by Adventure10

Solution with Aryan_23

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#9 h Jan 1, 2022, 8:11 AM

Y by

Let perpendicular from X to AY meet BD at S. we have to show S is the orthocenter so let's prove YS || BC instead.

∠ADC = ∠ACB ---> BC tangent to circle ADC ---> ∠DAC = ∠DCX
∠AXC = ∠AYC = 90 ---> ∠ACXY is cyclic.

step1 : AYD and AXC are similar. (1)
∠YAD = ∠YAX - ∠DAX = ∠YCX - ∠DAX = ∠DAC - ∠DAX = ∠XAC
∠AYD = 90 = ∠AXC

step2 : ADC and ACB are similar. (2)
∠ADC = ∠ACB
∠DAC = ∠CAB

step3 : DSY and DBC are similar. (3)
from (1) we have : YD/XC = AD/AC and from (2) we have : AD/AC = DC/CB
so YD/XC = DC/CB ---> DY/DC = XC/BC and from Thales theorem we have XC/BC = SD/BD.
we have DY/DC = SD/BD and ∠YDS = ∠CDB.

now from (3) we have SY || BC.
we're Done.

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#10 h Jan 7, 2022, 4:47 PM

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Let perpendicular to $AY$ through $X$ meets $BD$ at $H.$ Clearly $ABXC\sim ACYD,$ so $\frac{|XH|}{|CD|}=\frac{|BX|}{|BC|}=\frac{|CY|}{|CD|}.$ Thus $CXHY$ is parallelogram, and since $C$ is antipode of $A$ on $\odot (AXY),$ $H$ is orthocenter of $AXY.$ Done.

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