Equal angles in the pentagon and perpendicular lines

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Equal angles in the pentagon and perpendicular lines

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Source: Polish MO 2008 Second Round (1st day)

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TomciO

#1 h Feb 22, 2008, 7:46 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

In the convex pentagon $ABCDE$ following equalities holds: $\angle ABD= \angle ACE, \angle ACB=\angle ACD, \angle ADC=\angle ADE$ and $\angle ADB=\angle AEC$ . The point $S$ is the intersection of the segments $BD$ and $CE$ . Prove that lines $AS$ and $CD$ are perpendicular.

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#2 h Feb 22, 2008, 10:32 PM • 1 Y

Y by Adventure10

Let $X=AC\cap BS$ , $Y=AC\cap ES$ , $Z=AS\cap CD$ .

We see that $ABCS$ and $AEDS$ are cyclic, so
$\angle ASB=\angle ACB=\angle ZCX,$
so $XCZS$ is cyclic. Similar, $YDZS$ is cyclic. It thus follows that
$\angle AXS = \angle CZS = \angle DYS,$
so $AXSY$ is cyclic. Hence,
$\angle YDC = \angle ADC = \angle EDA = \angle ESA = \angle YSA = \angle YXA,$
so $XCDY$ is cyclic. It follows that
$\angle SXC = \angle DYS.$
But $\angle SXC = \angle SZD$ and $\angle DYS = \angle CZS$ , so
$\angle SZD = \angle CZS.$
Thus $\angle SZD = \angle CZS = 90^\circ$ .

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#3 h Feb 23, 2008, 2:33 AM • 1 Y

Y by Adventure10

reflect $A$ to $CD$ and apply Jacobi theorem.

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Ashegh

#4 h Mar 8, 2008, 10:17 AM • 2 Y

Y by Adventure10, Mango247

hmmmmmmmmmm,a good problem

$\angle ABD=\angle ACE=\alpha$ (1)

$\angle ACB=\angle ACD=\beta$ (2)

$\angle ADC=\angle ADE=\delta$ (3)

$\angle ADB=\angle AEC=\gamma$ (4)

according to (1)and(4),we find out $\triangle ABD$ and $\triangle ACE$ are similar to each other.

Infact $\triangle ABD$ turns into $\triangle ACE$ by a spiral similarity with radius $\frac{AD}{AB}$

And angle $180-(\beta+\gamma)$ ,

Then it shows that $\triangle ABC$ and $\triangle ADE$ ,are also similar to each other.And:

$\angle CBA=\angle CDA=\delta$

but it shows that: $\triangle ABC$ = $\triangle ADC$ ,then the points $D$ and $B$ ,turn into each other

with a reflect to segment $AC$ ,and we conclude that $AC$ is perpendicular to $BD$ .

With a similar way,we can show $AD$ is perpendicular $CE$ ,too.

$AC \cap BD=M$

$AD \cap CE=N$

then in triangle $ACD$ ,the altitudes are $DM$ and $CN$ .

And according to concurrency of the three altitudes of triangle,

$AS$ is also the third altitude of triangle $ACD$ ,and it means that: $AS \bot CD$

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#5 h Feb 25, 2017, 8:30 AM • 1 Y

Y by Adventure10

it is easy if we use similarity

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MagicalToaster53

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MagicalToaster53

#6 h Mar 13, 2024, 10:18 PM

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Let $X = AD \cap CE, Y = AC \cap BD, Z = AS \cap CD$ .

Observe trivially that $AEDS$ and $ABCS$ are cyclic. We then make the following observations:

Claim: $SYCZ$ and $SXDZ$ are cyclic.
Proof: We have that $\angle BCA = \angle BSA = \angle DSZ = \angle YCZ.$ A similar angle chase ensues for $SXDZ$ . $\square$

Claim: $AXSY$ and $DXYC$ are cyclic.
Proof: We have that
$\begin{align*} \angle CSY &= 180^{\circ} - (\angle BSA + \angle ASE) \\ &= 180^{\circ} - (\angle YCD + \angle XDZ) \\ &= \angle DAC \implies AXSY \text{ is cyclic.}\\ \end{align*}$ Hence we then have that $\angle AYX = \angle ASX = \angle ZDX \implies DXYC \text{ is cyclic. } \square$
Now observe that
$\begin{align*} \angle SYC &= \angle SYZ + \angle ZYC \\ &= \angle XCD + \angle ZSC \\ &= \angle XYD + \angle XSA \\ &= \angle XYD + \angle XYA \\ &= \angle SYA, \end{align*}$ so that as $\angle SYC + \angle SYA = 180^{\circ} \implies \angle SYC = 90^{\circ} \implies \angle SZC = 180^{\circ} - \angle SYC = 90^{\circ},$ as desired. $\blacksquare$

This post has been edited 1 time. Last edited by MagicalToaster53, Mar 14, 2024, 9:04 PM

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