TwoTimes3TimesSeven wrote: $ABCD$ is a convex quadrilateral. Angles $A$ and $C$ are equal. Points $M$ and $N$ are on the sides $AB$ and $BC$ such that $MN||AD$ and $MN=2AD$. Let $K$ be the midpoint of $MN$ and $H$ be the orthocenter of $\triangle ABC$. Prove that $HK$ is perpendicular to $CD$. The equivalent problem is: Let $ABC$ be a triangle and $D$ is on minor arc $BC$ of $\odot(ABC)$. Let $E$ be a point such that $ADBE$ is a parallelogram. Denote $H$ the orthocenter of $\triangle{ABC}$. Prove that $A,H,B,E$ are concyclic. But $\angle{AEB}=\angle{ADB}=\angle{ACB}=180^{\circ}-\angle{AHB}$. So we’re done.

fastlikearabbit wrote: The equivalent problem is: Let $ABC$ be a triangle and $D$ is on minor arc $BC$ of $\odot(ABC)$. Let $E$ be a point such that $ADBE$ is a parallelogram. Denote $H$ the orthocenter of $\triangle{ABC}$. Prove that $A,H,B,E$ are concyclic. But $\angle{AEB}=\angle{ADB}=\angle{ACB}=180^{\circ}-\angle{AHB}$. So we’re done. How are those problems equivalent? $ABCD$ is not cyclic? Am I missing something?

GGPiku wrote: fastlikearabbit wrote: The equivalent problem is: Let $ABC$ be a triangle and $D$ is on minor arc $BC$ of $\odot(ABC)$. Let $E$ be a point such that $ADBE$ is a parallelogram. Denote $H$ the orthocenter of $\triangle{ABC}$. Prove that $A,H,B,E$ are concyclic. But $\angle{AEB}=\angle{ADB}=\angle{ACB}=180^{\circ}-\angle{AHB}$. So we’re done. How are those problems equivalent? $ABCD$ is not cyclic? Am I missing something? It’s a completely different problem, and the notations are not taken from the initial problem. Look at the ‘equivalent problem’ as a key Lemma that can be used to solve the original problem.

TwoTimes3TimesSeven wrote: $ABCD$ is a convex quadrilateral. Angles $A$ and $C$ are equal. Points $M$ and $N$ are on the sides $AB$ and $BC$ such that $MN||AD$ and $MN=2AD$. Let $K$ be the midpoint of $MN$ and $H$ be the orthocenter of $\triangle ABC$. Prove that $HK$ is perpendicular to $CD$. Knowing that $MK=\frac{MN}{2}=AD$ and $AD \parallel MK$, we can conclude that $AMKD$ is a parallelogram. So, $\angle{MKD}=\angle{MAD}=\angle{NCD}$, from the given information. We obtained that $D,C,N,K$ are on the same circle. Take $H’$ as the orthocenter of $\triangle{DKC}$. I shall prove that $H’$ is also the orthocenter of triangle $ABC$ and with this fact proven, the problem will be solved. Using $AM \parallel DK$, we get $CH’ \perp AB$. So we are left to prove that $AH’ \perp BC$. Or, $\angle{AH’K}=180^{\circ}- \angle{DCB} $. But $\angle{DCN}=\angle{DKM}=180^{\circ}- \angle{ADK}.$ It will be enough to prove $\angle{AH’K}=\angle{ADK}$, which is equivalent with $ADH’K$ is cyclic. Observe that $AKND$ is also a parallelogram, so $\angle{KAD}=\angle{KND}=\angle{DCK}=180^{\circ}-\angle{DH’K}$, which gives us the conclusion.

Projective sledge-hammer TwoTimes3TimesSeven wrote: $ABCD$ is a convex quadrilateral. Angles $A$ and $C$ are equal. Points $M$ and $N$ are on the sides $AB$ and $BC$ such that $MN||AD$ and $MN=2AD$. Let $K$ be the midpoint of $MN$ and $H$ be the orthocenter of $\triangle ABC$. Prove that $HK$ is perpendicular to $CD$. Note that $ABCDH$ lie on the circumrectangular hyperbola $\mathcal{H}$ that is the isogonal conjugate of the perpendicular bisectors side $AC$ in $\triangle ABC$. Observe that $AMKD$ is a parallelogram hence $\overline{DK} \perp \overline{CH}$. Redefine point $K$ as the orthocenter of triangle $CHD$. Let $L$ lie on line $\overline{AC}$ such that $\overline{BL} \parallel \overline{AD}$. Let $K'$ be the point on $\mathcal{H}$ for which $(BA, BC, BK', BL)=-1$. Then we wish to show $K=K'$. Fix $\triangle ABC$ and hyperbola $\mathcal{H}$. Move point $D$ and since $K \in \mathcal{H}$ with $\overline{DK} \parallel \overline{AB}$ we see $D \mapsto K$ is projective. However $D \mapsto L \mapsto K'$ is also projective. Thus, it is enough to solve the problem for three choices of point $D$. For $D=B$ we obtain $K=A$ and $K'=L=A$. For $ABCD$ a parallelogram. Note $K'=L=C$ and $\angle HCD=90^{\circ}$ so $K=C$. For $ABCD$ cyclic. Note $\angle DAB=\angle DCB=90^{\circ}$. Observe that $\tfrac{\sin ABL}{\sin CBL}=\tfrac{1}{\cos A}$ and since $K \in \mathcal{H}$ we have $\angle BAK=\angle CAK$ so $$\frac{\sin ABK}{\sin CBK}=\frac{AK \cdot \sin BAK}{CK \cdot \sin BCK}=\frac{AK}{CK}=\frac{1}{\cos A}.$$The last line is true because $\triangle BHD \cong \triangle CDH$ and $\overline{BK}$ is the circumdiameter in $\triangle BHD$.

First, ignore the figure from the problem and restate/reconstruct it in this way: Quote: Given a isosceles trapezoid $DKNT$ with $KN\parallel DT$. Extend $NK$ past $K$ til $M$ that $MK=KN$, and draw parallelogram $KMAD$, i.e. note that $A,D,T$ collinear. Let $C$ be a point on minor arc $NC$ of the circumcircle of $KDNT$, and then let $AM\cap NC=B$. Show that the following lines are concurrent: The line $\ell_1$ through $A$ perpendicular to $BC$, The line $\ell_2$ through $C$ perpendicular to $AB$, and The line $\ell_3$ through $D$ perpendicular to $KC$. Let $AD\cap BC=E$. Let $\angle{CDE}=a,\angle{KAD}=b,\angle{MAK}=c$. Angle chasing gives us $\angle{AKD}=\angle{KDN}=\angle{KCN}=c$, $\angle{KCD}=b,\angle{CED}=b+c-a$. Let $\angle{DCA}=x$. Further angle chasing gives us $\angle{KAC}=b+x-a,\angle{CAD}=a-x$. Now, using Ceva's with triangle $ADC$ and cevians $AK,DK,CK$ gives us \begin{align*} & \frac{\sin (b)}{\sin (b+x-a)} \cdot \frac{\sin (b-x)}{\sin (b)}\cdot \frac{\sin (b+c-a)}{\sin (b+c)} =1\\ \implies & \sin (b-x)\sin (b+c-a) =\sin (b+x-a)\sin (b+c)\\ \implies & \cos(a-x-c)-\cos (2b-x+c-a)=\cos (a+b-x) -\cos (2b+x+c-a). \end{align*}We want to show that $\ell_1 ,\ell_2 ,\ell_3$ concurrent, Ceva's gives us it's enough to show that \begin{align*} & \frac{\cos (b+c-a)}{\cos (b+c-x)} \cdot \frac{\cos (b+c+x-a)}{\cos (b+c-a)} \cdot \frac{\cos (b)}{\cos (b+c+x-a)} =1\\ \iff & \cos(b+c+x-a) \cos (b)=\cos (b+c-x)\cos (b+c+x-a)\\ \iff & \cos (2b+x+c-a)+\cos (x+c-a)=\cos (a+b-x) +\cos (2b+c-x-a), \end{align*}which is true from the aforementioned equation, done.

Lemma. If $ABCD$ is a convex quadrilateral with $\angle A = \angle C$ and $H$ is the orthocenter of $BCD$, then $\angle HCD = \angle DAH$. Proof. Note that $\angle BHD = 180 - \angle C = 180 -\angle A$ and so $ABDH$ is cyclic. So $\angle DAH = \angle DBH = \angle HCD$. $\Box$ Now, in the main problem Let $H'$ be the orthocenter of $CDH$ and the line through $H'$ and parallel to $AD$ meet $AB, BC$ at $P, Q$. $DH'\perp CH\perp AB$, so $DH'|| AP$ and $ADHP$ is a parallelogram and $PH = AD$. $\angle DH'N = 180-\angle A = 180 -\angle C = \angle DCQ$, so $DH'CQ$ is cyclic and $\angle H'QD = \angle H'CD$ and from lemma we get $\angle H'CD = \angle DAH'$. Finally, $AD|| KQ$ and $\angle DAH' = \angle H'QD$, so $ADQH'$ is a parallelogram and $KN = AD = MK$, $P = M, Q=N, H' = K$, $HK\perp DC$. $\Box$

anantmudgal09 wrote: Projective sledge-hammer TwoTimes3TimesSeven wrote: $ABCD$ is a convex quadrilateral. Angles $A$ and $C$ are equal. Points $M$ and $N$ are on the sides $AB$ and $BC$ such that $MN||AD$ and $MN=2AD$. Let $K$ be the midpoint of $MN$ and $H$ be the orthocenter of $\triangle ABC$. Prove that $HK$ is perpendicular to $CD$. Note that $ABCDH$ lie on the circumrectangular hyperbola $\mathcal{H}$ that is the isogonal conjugate of the perpendicular bisectors side $AC$ in $\triangle ABC$. Observe that $AMKD$ is a parallelogram hence $\overline{DK} \perp \overline{CH}$. Redefine point $K$ as the orthocenter of triangle $CHD$. Let $L$ lie on line $\overline{AC}$ such that $\overline{BL} \parallel \overline{AD}$. Let $K'$ be the point on $\mathcal{H}$ for which $(BA, BC, BK', BL)=-1$. Then we wish to show $K=K'$. Fix $\triangle ABC$ and hyperbola $\mathcal{H}$. Move point $D$ and since $K \in \mathcal{H}$ with $\overline{DK} \parallel \overline{AB}$ we see $D \mapsto K$ is projective. However $D \mapsto L \mapsto K'$ is also projective. Thus, it is enough to solve the problem for three choices of point $D$. For $D=B$ we obtain $K=A$ and $K'=L=A$. For $ABCD$ a parallelogram. Note $K'=L=C$ and $\angle HCD=90^{\circ}$ so $K=C$. For $ABCD$ cyclic. Note $\angle DAB=\angle DCB=90^{\circ}$. Observe that $\tfrac{\sin ABL}{\sin CBL}=\tfrac{1}{\cos A}$ and since $K \in \mathcal{H}$ we have $\angle BAK=\angle CAK$ so $$\frac{\sin ABK}{\sin CBK}=\frac{AK \cdot \sin BAK}{CK \cdot \sin BCK}=\frac{AK}{CK}=\frac{1}{\cos A}.$$The last line is true because $\triangle BHD \cong \triangle CDH$ and $\overline{BK}$ is the circumdiameter in $\triangle BHD$. where can i find some materials about hyperbola topic?I have searched from google but none of them give me sufficient information to understand this solution.Can someone help me for that?

Construct H as orthocentre of DCK then CH|DK//AB. Note that A=C so DCN=DAN'=NKD so NCKD cyclic and KHC=KDC=KNC. From KHD=KCD =KND=KAD so AHKD cyclic and KHA=180-KDA=KMB then AHC=KMB+MNB=180-B. From this H is orthocentre of ABC, so KH|CD (q.e.d)

Sorry to bump this post, but can someone explain how proving $AH’ \perp BC$ is related to proving $\angle{AH’K}=180^{\circ}- \angle{DCB} $ in fastlikearabbit's post. Also, can someone explain the circumrectangular hyperbola along with most of anantmudgal09's post. Thanks.

Very nice problem! Let $K'$ and $C'$ be the orthocenter of $\triangle{CHD}$ and the reflection of $C$ in the midpoint of $CK'$. It is enough to show that $K' \equiv K$. $$\measuredangle{HAD}=\measuredangle{BAD}-\measuredangle{BAH}=\measuredangle{DCB}-\measuredangle{HCB}=\measuredangle{HCD} \implies \measuredangle{HAD}=\measuredangle{HKD}$$This implies that $HAC'DK'$ lie on a circle with diameter $HC'$. Let $A'$ be a point such that $AC'A'C$ and $ADA'K$ are parallelograms, because $C'A\perp HA$ we have that $A'$ lies on $BC$, thus $DAM'K$ is also a parallelogram, where $M' = A'K'\cap AB$, which implies that $K' \equiv K$.

Redefine $K$ as the orthocenter of $\triangle CHD$. We also redefine $M,N$ as points on $BA, BC$ such that $K$ is the midpoint of $MN$. Note that this immediately implies the problem, so we will prove that $MN\parallel AD$ and $MN=2AD$. The former statement implies the latter since we already have that $KD\perp CH\perp AB$. Let us restrict the attention to proving that $MN\parallel AD$. Notice that $B, A, C, H, D$ lie on the isogonal conjugate (w.r.t. $\triangle ABC$) of the perpendicular bisector of $AC$, which is a hyperbola $\mathcal{H}$; this hyperbola is rectangular so $K$ also lies on this hyperbola. Construct $B'$ such that $BAB'C$ is a parallelogram; clearly $B'\in\mathcal{H}$. Now, we project \begin{align*} B(A,C; B',K) &= (A,C;B',K)_\mathcal{H} \\ &= H(A,C;B',K) \\ &= C(B,B';\infty_{\perp HB'}, D) &\text{(Rotate }90^\circ)\\ &= (B,B';C,D) \\ &= A(B,B';C,D) \\ &= B(A,C;\infty_{AC},\infty_{AD}) &\text{(Translate)} \end{align*}(Here we have used the fact that the line through $C$ perpendicular to $HB'$ is tangent to $\mathcal{H}$. To see why, notice that if this line meet $\mathcal{H}$ again at $C'$, then $\angle ABC' = \angle ACC' = \angle B$ or $C'=C$, contradiction.) The above cross-ratio chasing gives a projective map that fixes both $BA$ and $BC$, while mapping $BB'\mapsto B\infty_{AC}$ and $BK\mapsto B\infty_{AD}$. The first two pairs gives that this map is a harmonic conjugate w.r.t. $BA, BC$. Hence, $$B(A,C;K,\infty_{AD})=-1 \implies MN\parallel AD$$as desired.

Redefine $H$ to be the orthocenter of $\triangle CDK$. We show that $H$ is the orthocenter of $ABC$ too. Note that since $AMKD$ is a parallelogram, we have $AM \parallel KD$, so $CH \perp AB$. Now we'll prove $AH \perp BC$. Note that $\angle NCD = \angle MAD = \angle MKD$, so $C,K,N,D$ lie on a circle. Let $C'$ denote the $C$ antipode in that circle. Since $CK \perp KC'$ and $CK \perp HD$, so $KC' \parallel HD$. Also $HK \perp CD$ and $DC'\perp DC$, so $HKC'D$ is a parallelogram. Since $AKND$ is a parallelogram too, we have $AHNC'$ is a parallelogram too. Hence since $NC' \perp CN$, we have $AH \perp CN$ as well. We're done. $\square$

It is clear that $MKDA$ and $KNDA$ are parallelograms. Now let's define $H'$ as the orthocenter of triangle $KCD$. We would like to prove that $H'\equiv H$, which is enough to imply the wanted result. Both $H$ and $H'$ lie on the same altitude from $C$ to $AB$ and $KD$, which are parallel, so in turn it is enough to prove that $\angle AH'C=\angle AHC= 180 - \angle ABC$. We know that $$\angle DKN +\angle DCN= 180 - \angle BAD + \angle BCD = 180$$, thus $CKDN $ is cyclic. We also have that $$\angle KH'D + \angle KAD = 180 - \angle KCD +\angle KAD = 180$$thus $AKH'D$ is cyclic as well. Now we get that $$\angle AH'C=\angle AH'D +\angle DH'C= \angle AKD + 180 - \angle DKC= 180 + \angle MKD - \angle MKA - (\angle DKN - \angle CKN)=$$$$180 + \angle MKD - \angle KND - ((180 -\angle BAD)-(\angle MNB - \angle KCN )=$$$$=\angle MKD - \angle KND + \angle BAD + (180 - \angle NMB - \angle MBN - \angle KDN ) =$$$$= 180 - \angle ABC = \angle AHC$$and we're done!

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Define $H$ as orthocenter of triangle $KCD$. Note that $AMKD$ and $AKND$ are parallelograms. We have $CH \perp DK \parallel AB$, so it's enough to prove that $AH \perp CN$ (then $H$ is orthocenter of triangle $ABC$). Note that $\angle NCD = \angle MAD = \angle DKM$, so $DCNK$ is cyclic. Now delete points $M$ and $B$ and complex bash. Let $(DCNK)$ be a unit circle. Then $a=k+d-n, h=k+c+d$. We have $h-a=c+n$ and so $AH \perp CN \Leftarrow 0=(h-a)\overline{(c-n)}+\overline{(h-a)}(c-n)=(c+n)\overline{(c-n)}+\overline{(c+n)}(c-n)$, which is obvious.

Nice Note that $AD = MK$ so $ADKM$ is parallelogram. $CH \perp AB \parallel DK$ so we need to prove $H$ is orthocenter of $CKD$. Note that since $DK \perp CH$ we only need to prove $\angle DHK = \angle DCK$. Claim $1: CNKD$ is cyclic. Proof $:$ Note that $\angle DKN = 180 - \angle DKM = 180 - \angle A = 180 - \angle C$. Let $H'$ be reflection of $H$ w.r.t midpoint of $KD$. We need to prove $H'$ lies on $CNKD$. Note that $H'D = HK$ and $DA = KN$ and $\angle HKN = \angle H'DA$ so $ADH'$ and $NKH$ are congruent and so $AHNH'$ is also parallelogram so $NH' \parallel AH \perp NC$ so $\angle CNH' = 90$. Now assume $S$ is orthocenter of $CKD$ and $S'$ is reflection of $S$ w.r.t midpoint of $DK$. Since $\angle DS'K = DSK = ACK$, $S'$ lies on $CNKD$ and since $CK \perp DS \parallel KS'$ we have that $\angle CNS' = 90 = \angle CNH'$. Since $S'H' \parallel CH$ and $CH$ is not perpendicular to $NC$ so $S'$ and $H'$ are same so $S$ and $H$ are same so $H$ is orthocenter of $CKD$ as wanted.

Synthetic solution (Angle-chase): Re-define $H$ to be the orthocenter of $\triangle CKD$. Then: evidently $CH \perp AB$ as $AB \parallel CK$. It suffice to show that $AH \perp BC$. Notice that: $$\angle DAM = \angle CKM = \angle DCN$$which implies $CNKD$ to be cyclic. Consider $CC'$ to be the diameter of $(CNKD)$. Notice that: $HDC'K$ and $AKND$ are parallelograms which implies $ADNH$ to be a parallelogram. Since $C'N \perp BC$, we have $AH \perp BC$. Hence proved.