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#1 h Jul 22, 2019, 3:32 AM • 1 Y

Y by Adventure10

Let $ABC$ be a right angled triangle with $\angle A = 90^o$ and $AD$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $AB$ and $AC$ respectively. The parallel line from $C$ to $EZ$ intersects the line $AB$ at the point $N$ . Let $A'$ be the symmetric of $A$ with respect to the line $EZ$ and $I, K$ the projections of $A'$ onto $AB$ and $AC$ respectively. If $T$ is the point of intersection of the lines $IK$ and $DE$ , prove that $\angle NA'T = \angle ADT$ .

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#2 h Aug 31, 2019, 8:33 PM • 2 Y

Y by itslumi, Adventure10

First, let us denote $\angle B=\alpha, AD\cap EZ={P}, IK\cap AA'={R}$ and $AA'\cap DE={Q}$ . Since $DE||AC$ , we have $\angle DEA=90^o$ . Similarly, $\angle DZA=90^o$ .

Now, since $\angle ADB=90^o$ ,
we have $\angle DAB = 180^o-\angle ADB -\angle ABD=180^o-90^o-\alpha=90^o-\alpha$ .
From here, $\angle ADE=180^o-\angle DEA-\angle DAE=180^o-90^o-(90^o - \alpha) = \alpha$ .
Because $AZDE$ is a rectangle, $\angle ADE=\angle AZE \equiv \angle AZR$ .

$ZE$ and $CN$ are parallel, so $\angle AZE=\angle ACN=\alpha$ .
From here we derive
$\angle CNA=90^o-\angle ACN=90^o-\alpha$
and
$\angle INC=180^o-\angle CNA=90^o+\alpha$

$A'$ is symmetric of $A$ with respect to $ZE$ , so we must have $AA'\perp ZE$ . Thus,
$\angle ZAR=90^o-\angle AZR=90^o-\alpha$ .
Now,
$\angle AA'K=90^o-\angle KAA' \equiv 90^o-\angle ZAR=\alpha$
Because $AKA'I$ is a rectangle, we get $\angle AA'K=\angle AIK=\alpha$
So,
$\angle ETI=90^o-\angle EIT \equiv 90^o-\angle AIK=90^o-\alpha$
and
$\angle ITQ=180^o-\angle ETI=90^o+\alpha$

Lemma 1: $CN, AA'$ and $DE$ are concurrent
Proof:
1) Let us suppose a line $p$ through $Q$ , such that $p\perp AA'$ . Let $C'$ be the intersection of $p$ with $AC$ . Since $\angle CAA'=90^o-\alpha$ , and therefore $\angle C'AA'=90^o-\alpha$ ,
$\angle QC'A=\alpha$
Because of this, and because $\angle ADQ \equiv \angle ADE=\alpha$ , $\angle QC'A=\angle QDA, C'AQD$ is cyclic, a.e. the excircle of $\triangle AQD$ contains $C'$ .

2) We have $\angle A'AI \equiv \angle QAE=\alpha$ . Since $\angle DEA \equiv \angle QEA=90^o$ , we obtain
$\angle DQA=\angle QAE+\angle QEA=90^o+\alpha$
By definition, $\angle ACB \equiv \angle DCA=90^o-\alpha$ .
Now we see that $\angle DCA+\angle DQA=90^o-\alpha+90^o+\alpha=180^o$ , so the excircle of $\triangle AQD$ must contain $C$ .

Combining 1) and 2), we see that the excircle of $\triangle AQD$ intersects the line $AC$ at three points: $A, C$ and $C'$ . It is impossible for all three to be distinct points, since then the excircle of $\triangle AQD$ will have three common points with $AC$ , which is impossible. It follows that two of $A, C$ and $C'$ must be the same point. $A$ and $C'$ cannot be equivalent because, in that case $QC$ ' couldn't be perpendicular to $AA'$ . So $C\equiv C'$ .

Now, $CQ$ is perpendicular to $AA'$ , which means that $CQ\parallel ZE$ , and that $CQ$ contains $N$ . Thus, the proof of the lemma is complete.

Lemma 2: $A'QNI$ is cyclic
Proof:
We have $\angle A'QN=90^o$ and $\angle A'IN=90^o$ .
Case 1: $N$ is between $A$ and $I$
-- $\angle A'IN+\angle A'QN=180^o$ , so the lemma is true.
Case 2: $N$ is between $B$ and $I$
-- $\angle A'QN=\angle A'IN=90^o$ , so the lemma is true.

So, $\angle CNI \equiv \angle QNI=\angle DTI \equiv \angle QTI$ , or $\angle QTI=\angle QNI$ . From here we obtain that $QTNI$ is cyclic. Because $QNIA'$ is also cyclic, $Q, T, N, I$ and $A'$ all lie on the same circle.

*Note: When $N$ is between $I$ and $B$ , the proof is done by using the fact that $\angle QNI \equiv\angle CNA=90^o -\alpha$

Case 1: $N$ is between $A$ and $I$
-- $\angle AIK \equiv \angle NIT=\alpha$ , and by peripheral angles, we obtain $\angle NA'T=\angle NIT=\alpha$ , a.e. $\angle NA'T=\angle ADT=\alpha$ . Thus, the proof is complete.

Case 2: $N$ is between $B$ and $I$
-- $\angle NA'T=180^o-\angle TIN=\alpha$ , so $\angle NA'T=\alpha=\angle ADT$ . Thus, the proof is complete.

*Note: When $N$ and $I$ are equivalent, the proof follows directly from the fact that $QNIA'$ is cyclic and $\angle CQD=\alpha$ .

This post has been edited 1 time. Last edited by Math00954, Sep 5, 2020, 3:10 PM

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#3 h Jul 21, 2020, 11:09 PM

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Let $ED$ and $AA'$ meet $CN$ at $X_1$ and $X_2$ respectively. Note that $\frac{NX_1}{X_1C} = \frac{NE}{EA} = \frac{CZ}{ZA} = \frac{CD}{DB} = \frac{CA^2}{AB^2} = \frac{NA^2}{AC^2} = \frac{NX_2}{X_2C}.$ Thus $X_1=X_2 := X$ . Easily check that $TIA'X$ is an isosceles trapezoid. Since $\angle A'XN=90^{\circ} = \angle NIA'$ then $NIA'X$ is cyclic. We conclue that $NTIA'X$ is cyclic and thus $\angle NA'T = \angle NIT = \angle ADT$ .

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