goldeneagle wrote: in acute triangle $ABC$ angle $B$ is greater than$C$. let $M$ is midpoint of $BC$ . $D$ and $E$ are the feet of the altitude from$C$ and $B$ respectively .$K$ and $L$ are midpoint of $ME$ and $MD$ respectively.if $KL$ intersect the line through $A$ parall to $BC$ in $T$ prove that $TA$=$TM$ Draw the circumcircle of $ADE$ , then it`s easy to see $ME,MD$ and the line through $A$ parallel to $BC$ are tangent to it . Now $T$ lies on the radical axis of point $M$ (look it like a degenerate circle) and circumcircle of $ADE$ . So it has equal power WRT them ,which means $TM=TA$ .

mahanmath wrote: goldeneagle wrote: in acute triangle $ABC$ angle $B$ is greater than$C$. let $M$ is midpoint of $BC$ . $D$ and $E$ are the feet of the altitude from$C$ and $B$ respectively .$K$ and $L$ are midpoint of $ME$ and $MD$ respectively.if $KL$ intersect the line through $A$ parall to $BC$ in $T$ prove that $TA$=$TM$ Draw the circumcircle of $ADE$ , then it`s easy to see $ME,MD$ and the line through $A$ parallel to $BC$ are tangent to it . Now $T$ lies on the radical axis of point $M$ (look it like a degenerate circle) and circumcircle of $ADE$ . So it has equal power WRT them ,which means $TM=TA$ . Can you explain more? Why $ME,MD$ are tangent and how you find that conclustion?

using an easy angle-chasing implies $ME$ and $MD$ to be tangent to circumcircle of $ADE$. I think what mahanmath wrote is obvious and clear.

Very nice Problem! A hint for another solution. Let $U\equiv(ME)\cap(AT)$ and $V\equiv(TA)\cap(MD)$. Denote $I$ the circumcenter of $\triangle{ADE}$, $T_1\equiv(DE)\cap(UV)$ , $M_1$ the midpoint of $MF$, $M_2\equiv(IT_1)\cap(MF)$ , $M_3\equiv(DE)\cap(MF)$ and $M_4\equiv(MF)\cap(KL)$. See that $(I)$ is the incircle of $\triangle{MUV}$ thereafter, deduce that $(IT_1)\perp(MF)$, So it's left to prove that $(T_1M_2)\parallel(TM_1)$ which can be reduced to $\frac{FM_2}{FM_1}=\frac{FM_3}{FM_4}$.

Alternative proof. With the same notations as the previous one. Let $M_5\equiv (IM)\cap(DE)$ and $M_6\equiv(IM)\cap(TM_4)$. It's Obvious that $M_6$ is the midpoint of $MM_5$ and that $(TM_6)\perp(MM_5)$, Hence $TM=TM_5$. Now, let $M_7$ , $M_8$ and $M_9$ be midpoints of $IT_1$ , $MT_1$ and $FT_1$ respectively. By an angle chasing, we get that $\angle{M_8TM_9}+\angle{M_8M_7M_9}=180$ and then the quadrilateral $M_7M_8TM_9$ is cyclic. Furthermore, triangles $\triangle{M_8M_7M_9}$ and $\triangle{MIF}$ are homothetic with respect to $T_1$ hence, $\angle{M_7TT_1}=\angle{M_7M_8M_9}=\angle{IMM_2}=\angle{T_1M_5M_7}$, so points $T$ , $T_1$ , $M_5$ and $M_7$ are concyclic; It follows that $\angle{M_5TF}=\angle{M_5M_7I}=180-2\angle{M_5FT}$ because $M_5$ is the center of $\odot(FIM_5T_1)$. Thus $TM_5=TA$, and so $TA=TM$.

Since $CD \perp AB$ and $BE \perp AC$, quadrilateral $BCDE$ is cyclic with center $M$. This implies that $\angle{TAE}=\angle{ACB}=\angle{ADE}$, since $TA \| BC$, and therefore that $TA$ is tangent to the circumcircle $\omega$ of $\triangle{ADE}$. Also, $\angle{DME}=2\angle{ABE}=180^\circ - 2\angle{BAC}$. This implies that $\angle{MDE}=\angle{MED}=\angle{BAC}$ and hence that $MD$ and $ME$ are tangent to $\omega$. Therefore the line $KL$ is the radical axis of circle $\omega$ and point $M$ and the powers of point $T$ with respect to $M$ and $\omega$ are equal. Since $TA$ is tangent to $\omega$, $TA=TM$.

Let $DE$ meet $BC$ at $F$ and meet $AT$ at $Z$,meet $AM$ at $U$ ,$CD$ meet $BE$ at $H$; $FH$ meet $AM$ at $Y$. $X$ on $AM$ such that TX is perpendicular with $AM$ Easy to see $FH$ is perpendicular with $AM$. And: $AX=\frac{1}{2} AM$ similar $\frac{AX}{YM}=\frac{AM}{2YM}$ or $\frac{AU}{UM}=\frac{ZA}{FM}=\frac{AY}{2.YM}$(*) Note $(AYUM)=-1$ then $\frac{2}{UM}=\frac{1}{MY}+­\frac{1}{MA}$ and $\frac{AU}{MA}=\frac{YU}{YM}$ Hence (*) is true and we are done.

mahanmath wrote: goldeneagle wrote: in acute triangle $ABC$ angle $B$ is greater than$C$. let $M$ is midpoint of $BC$ . $D$ and $E$ are the feet of the altitude from$C$ and $B$ respectively .$K$ and $L$ are midpoint of $ME$ and $MD$ respectively.if $KL$ intersect the line through $A$ parall to $BC$ in $T$ prove that $TA$=$TM$ Draw the circumcircle of $ADE$ , then it`s easy to see $ME,MD$ and the line through $A$ parallel to $BC$ are tangent to it . Now $T$ lies on the radical axis of point $M$ (look it like a degenerate circle) and circumcircle of $ADE$ . So it has equal power WRT them ,which means $TM=TA$ . Perfect

Dear Friends I also solved it by Edgar Cayce theorem. assume the $\bigodot ADHE$ is $w_{1}$ that here $H$ is the orthocenter of triangle $ABC$, and $\bigodot BDEC$ is $w_{2}$. Now it's clear that $P_{T}^{w_{1}} > P_{T}^{w_{2}}$ . by Cayce theorem we get that$ P_{T}^{w_{1}} - P_{T}^{w_{2}}= 2TK.PM$, that $TK$ is the length of the perpendicular from $T$ to $DE$ and $P$ is the midpoint of $AH$.(the center of $w_{1}$ ).now it's easy to get that $2TK.PM = \frac{a^{2}}{4}$. so what we get is $TA^{2} - TM^{2} + \frac{a^{2}}{4} = \frac{a^{2}}{4}$. now we can see it's Done. With Regards.

Suppose projection of $L,T$ onto $BC$ be $L',T'$ and projection of $L$ onto $TT'$ be $L''$.Now $\angle{TLL''}=A+2B-\pi$, now also $TL''=bSin C-\frac{a}{2Cos B}$.Now $CT=-TL'Cot(A+2B)$.Now it's enough to show $\frac {2CT}{a}=\frac {Sin(A-\theta)}{Sin(A-2\theta)}$ where $\angle{MAC}=\theta$ also we've $\frac {Sin(\theta)}{Sin(A-\theta)}=\frac {c}{b}$ and from thiose two equations we get $\frac {-bSin C-\frac{a}{2Cos B}Cot(A+2B)}{a}=\frac {Sin(A-\theta)}{Sin(A-2\theta)}$ and so that implies $\angle{CTM}=(A-2\theta )\implies \angle{ATM}=\pi-2(C+\theta)$ and we had $\angle{MAT}=C+\theta$ thus finally we get $AT=TM$ and so done.

Lemma :For this configuration let $N$ be a midpoint of $DE$. Let $O$ be a center of circumcircle of $\Delta ANM$. Then $OA \parallel BC$. Proof-it's just easy angle chasing. Let we see that $KL$ is a perpendicular bisector of the segment $MN$, since $DE \perp NM$ and $DE \parallel KL$, so $T$ is the center of circumcircle of $\Delta ANM$, and this implies $MT=MA$.

Notice that because $BE \perp CE$ and $CD \perp BD$, the points $B, C, D, E$ are inscribed in a circle of diameter $\overline{BC}.$ Hence, $M$ is the center of this circle, so $\triangle MDE$ is isoceles. Now, let $F$ be the midpoint of $\overline{DE}.$ Then since $\triangle MDE$ is isoceles, we deduce that $KL$ is the perpendicular bisector of $\overline{MF}.$ Now here is the key to the whole solution: The points $K, L, T$ are really annoying, and we would love to get them out of the picture. We can do this by proving that the circumcenter of $\triangle AFM$ lies on the line through $A$ parallel to $BC.$ This is because then said circumcenter must be the intersection of the perpendicular bisector of $\overline{MF}$ and the line through $A$ parallel to $BC$, i.e. the circumcenter must be $T$, whence the desired result will follow immediately. Now the problem is reduced to an angle chase: Let $T'$ be the circumcenter of $\triangle AFM.$ Then since $F$ lies strictly between $A$ and $M$ on $\odot(AFM)$, we deduce that \[\angle T'AM = 90^{\circ} - \frac{1}{2}\angle AT'M = 90^{\circ} - \left(180^{\circ} - \angle AFM\right) = \angle AFM - 90^{\circ} = \angle AFM - \angle EFM = \angle AFE.\] Finally, notice that since $BDEC$ is a cyclic quadrilateral, we have $\triangle AED \sim \triangle ABC.$ Hence, because $F$ and $M$ are "corresponding points" in these triangles (they are both the midpoints of corresponding sides), it follows that $\angle AFE = \angle AMB.$ Therefore $\angle T'AM = \angle AMB$, so $T'$ lies on the line through $A$ parallel to $BC$, as desired. $\square$

And just for fun...

This problem uses a neat, nice lemma.

$AD$ be the third altitude. Set $D=(0,0) , A=(0,a) , B=(b,0) , C=(c,0)$ Now compute, $E =(\frac{c(a^2+bc)}{a^2+c^2}, \frac{(c-b)ac}{a^2+c^2})$ $F=(\frac{b(a^2+bc)}{a^2+b^2}, \frac{(b-c)ab}{a^2+b^2})$ $K=(k_x,k_y)=(\frac{b+c}{4} + \frac{c(a^2+bc)}{2(a^2+c^2)} , \frac{ (c-b)ac }{ 2(a^2+c^2) } )$ Let $T=(t,a)$ we need to show that, $t^2=(t-(b+c)/2)^2+a^2$ Or, $t=\frac{a^2}{b+c}+\frac{b+c}{4}$ $\dots\color{blue}{ (1)}$ Now, Slope of $LK$ =slope of $EF =\frac{a(b+c)}{a^2-bc}$ Here we have used $c(a^2+b^2)+b(a^2+c^2)=(b+c)(a^2+bc)$$\dots(2)$ $c(a^2+b^2)-b(a^2+c^2)=(c-b)(a^2-bc)$.$\dots(3)$ Since $T$ lies on $LK$, $\frac{a-k_y}{t-k_x}=\frac{a(b+c)}{a^2-bc}$ Therefore, $t-k_x=\frac{a^2-bc}{a(b+c)} [a-\frac{(c-b)ac}{2(a^2+c^2)}]$ $=\frac{a^2}{b+c}-\frac{bc}{b+c}-\frac{(a^2-bc)(c-b)c}{2(b+c)(a^2+c^2)}$ $=\frac{a^2}{b+c}-\frac{c[2b(a^2+c^2)+c(a^2+b^2)-b(a^2+c^2)]}{2(b+c)(a^2+c^2)}$ [Using (3) again] $=\frac{a^2}{b+c}-\frac{c(a^2+bc)}{2(a^2+c^2)}$ [Using (2) again] $=\frac{a^2}{b+c} +\frac{b+c}{4} -k_x$ [From expression of $k_x$] This gives $\color{blue}{ (1)}$ Hence proved

SCP wrote: mahanmath wrote: goldeneagle wrote: in acute triangle $ABC$ angle $B$ is greater than$C$. let $M$ is midpoint of $BC$ . $D$ and $E$ are the feet of the altitude from$C$ and $B$ respectively .$K$ and $L$ are midpoint of $ME$ and $MD$ respectively.if $KL$ intersect the line through $A$ parall to $BC$ in $T$ prove that $TA$=$TM$ Draw the circumcircle of $ADE$ , then it`s easy to see $ME,MD$ and the line through $A$ parallel to $BC$ are tangent to it . Now $T$ lies on the radical axis of point $M$ (look it like a degenerate circle) and circumcircle of $ADE$ . So it has equal power WRT them ,which means $TM=TA$ . Can you explain more? Why $ME,MD$ are tangent and how you find that conclustion? Take the nine point circle of triangle ABC. From there you have angle MEX is 90 degrees, which implies ME is tangent.

mahanmath wrote: goldeneagle wrote: in acute triangle $ABC$ angle $B$ is greater than$C$. let $M$ is midpoint of $BC$ . $D$ and $E$ are the feet of the altitude from$C$ and $B$ respectively .$K$ and $L$ are midpoint of $ME$ and $MD$ respectively.if $KL$ intersect the line through $A$ parall to $BC$ in $T$ prove that $TA$=$TM$ Draw the circumcircle of $ADE$ , then it`s easy to see $ME,MD$ and the line through $A$ parallel to $BC$ are tangent to it . Now $T$ lies on the radical axis of point $M$ (look it like a degenerate circle) and circumcircle of $ADE$ . So it has equal power WRT them ,which means $TM=TA$ . Just to clarify, KL is perpendicular to XM. So it suffices to prove that the intersection of QM and KL has equal power between (AED) and point M. At that point, you can just length bash, and it all works out in the end

The coordinate bash is not very hard.

It's well-known that $MD$, $ME$, $AT$ are all tangent to $(ADE)$; see chapter 1 of the EGMO textbook, ``three tangents'' lemma. [asy][asy] pair A = dir(130); pair B = dir(210); pair C = dir(330); pair M = midpoint(B--C); pair D = foot(C, A, B); pair E = foot(B, A, C); pair K = midpoint(E--M); pair L = midpoint(D--M); filldraw(A--B--C--cycle, invisible, blue); pair T = extension(K, L, A, A+C-B); draw(T--A, lightred); draw(L--T, lightred); draw(M--T, lightred); filldraw(circumcircle(A, D, E), invisible, orange); draw(D--M--E, orange); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$D$", D, dir(220)); dot("$E$", E, dir(45)); dot("$K$", K, dir(315)); dot("$L$", L, dir(L)); dot("$T$", T, dir(T)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A = dir 130 B = dir 210 C = dir 330 M = midpoint B--C D 220 = foot C A B E 45 = foot B A C K 315 = midpoint E--M L = midpoint D--M A--B--C--cycle / 0.1 yellow / blue T = extension K L A A+C-B T--A / lightred L--T / lightred M--T / lightred circumcircle A D E / 0.1 red / orange D--M--E orange */ [/asy][/asy] Now line $KL$ is the radical axis of $(AED)$ and the circle centered at $M$ of radius zero. So by power of a point, \[ TM^2 = \operatorname{Pow}_{(AED)}(T) = TA^2. \]

Out of spite, I found a solution that doesn't use radical axis. Redefine $D$, $E$, and $F$ as the feet of the altitudes from $A$, $B$, and $C$ to the opposite sides. Let $N$ be the midpoint of $\overline{EF}$. Redefine $T$ as the circumcenter of $AMN$. Since $\overline{KL}$ is the perpendicular bisector of $\overline{MN}$, $T$ lies on $\overline{KL}$. Thus, it suffices to prove that $\overline{AT} \parallel \overline{BC}$. We have $\angle TAM=90^\circ-\tfrac{\angle ATM}{2}=\angle ANM-90^\circ=\angle ANE$. Since $\triangle ABC \sim \triangle AEF$, we have $ABMC \sim \triangle AEMF$, so $\angle ANE=\angle AMB$. Therefore, we have $\angle TAM=\angle AMB$, so $\overline{AT} \parallel \overline{BC}$.

Surprising twist to this old problem --- this is HM configuration! goldeneagle wrote: In acute triangle $ABC$ angle $B$ is greater than$C$. Let $M$ is midpoint of $BC$. $D$ and $E$ are the feet of the altitude from $C$ and $B$ respectively. $K$ and $L$ are midpoint of $ME$ and $MD$ respectively. If $KL$ intersect the line through $A$ parallel to $BC$ in $T$, prove that $TA=TM$. Let $U$ be the midpoint of $DE$; let $A \ne V=AM \cap (ADE)$, and $F$ be the point where $A$-symmedian meets $(ABC)$. It suffices to show that circumcentre of $\triangle AUM$ lies on $A$-parallel to $BC$. Apply inversion at $A$ of radius $\sqrt{AD \cdot AB}$; this swaps $\{D, B\}, \{E, C\}, \{U, F\}, \{V, M\}$; so the proposition becomes that line $VF$ is perpendicular to $BC$. However, this simply follows as $F, V$ are reflections in $BC$, as $V$ is the $A$-HM point.

Standard PoP solve It is well known that $TA$, $EM$ and $FM$ are tangent to $(AEF)$. Take $\omega$ as the circle with radius $0$ at $M$. Then, since $KE = KM$ and $FL = LM$, it follows that $KL$ is the radical axis of $\omega$ and $A$. Since $T$ lies on $KL$, as a result \[ TA = \text{Pow}_{(AEF)}(T) = \text{Pow}_{\omega}(T) = TM \]

taking radius 0 so fun cuz then its just length cuz length to any point is always tangent lool It can be easily proved that 1. ME=MF (note right triangles and median, so MF=ME. Tangents work, and no other configs are possible bc MEF is isosceles yet they always need to equal 1/2BC fixed and must lie on AB, etc.); 2. TA is a tangent (perp. AH diameter); Now, construct a circle of radius 0 on circle M, called m and call the other circle w. We have LF^2=LM^2; KE^2=KM^2; hence KL and consequently K is on the radical axis of m and w. Hence we're done since TM^2=TA^2. $\blacksquare$

The desired is a direct consequence of the Three Tangents Lemma, which states that $ME$, $MF$, and the line through $A$ parallel to $BC$ are all tangent to $(AEF)$. [asy][asy] size(200); defaultpen(linewidth(0.4)+fontsize(10)); pair A, B, C, E, F, H, M, K, L, T; A = dir(120); B = dir(200); C = dir(340); H = orthocenter(A, B, C); E = foot(H, C, A); F = foot(H, A, B); M = .5B + .5C; K = .5M + .5E; L = .5M + .5F; T = extension(K, L, A-.75, A+.75); draw(A--B--C--cycle); draw(circumcircle(A, E, F), blue); draw(E--M--F); draw(B--E); draw(C--F); ; draw(L--T); draw(A--T--M, heavyred); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$E$", E, dir(75)); label("$F$", F, W); label("$M$", M, S); label("$L$", L, S); label("$K$", K, NE); label("$T$", T, NE); dot(A); dot(B); dot(C); dot(E); dot(F); dot(M, blue); dot(K); dot(L); [/asy][/asy] Due to the tangencies, it follows that \[pow(K, (AEF)) = \frac{KE^2}{4} = \frac{KM^2}{4} = pow(K, \odot M)\]and analogously for $L$, so $KL$ is the radical axis of $(AEF)$ and the circle with radius $0$ centered at $M$. As $T$ lies on $KL$, we have \[pow(T, (AEF)) = pow(T, \odot M) \implies TA = TM. \text{ } \blacksquare\]

With Three tangents Lemma and Degenerate circiles it is obvious.

Note the Three Tangents Lemma shows that $\overline{ME}$, $\overline{MF}$ and $\overline{TA}$ are tangent to $(AEF)$. Consider $M = \omega$ as a circle with radius $0$. Notice that the radical axis of $\omega$ and $(AEF)$ is $\overline{KL}$. Hence, \[TA^2 = \operatorname{Pow}_{(AEF)} (T) = \operatorname{Pow}_{\omega} (T) = TM^2-R^2 = TM^2. \ \square\]

It's well known that $ME$, $MF$ and $AT$ are tangents to $(AEF)$. Then construct circle $M$ with radius $0$. We have $LF^2$ being the power of $L$ wrt $(AEF)$ but $LF^2 = LM^2$ so $L$ lies on the radical axis of $M$ and $(AEF)$ and so does $K$, similarly. Then $T$ lies on the radical axis of $M$ and $(AEF)$ so $TA^2 = TM^2$ finishes.

wow is this cool. Denote by $G$ the circle $(AED)$, and by $g$ the point circle at $M$. Observe that due to the three tangents lemma (EGMO Chapter 1), $ME, MD, AT$ are tangent to $G$. But now notice that $KL$ is the radax of $G, g$, so $Pow_G(T) = Pow_g(T) \iff TA^2 = TM^2 \iff TA = TM$. $\square$

$ $

$\angle C=\angle ADE=\angle CAT$ implies $TA$ is tangent to $(ADHE)$. On the other hand let $F$ be the altitude of $A$. Then $$\angle BCD=\angle MFD=\angle BAF$$thus, $MD$ and $ME$ are tangents to $ADHE$. For the point circle of $M$ with zero diameter, $KL$ is the radical axis of $(ADHE)$ and $M$. Thus, for tangents $TA$ and $TM$, we conclude as $T$ is being on $KL$ radical axis implies $TA=TM$ as desired.

By the well known and easily proven three tangents lemma, $MD$ and $ME$ are tangent to $(ADE)$. But then $L$ and $K$ have equal powers with respect to this circle and the zero circle at point $M$ from the midpoints. Hence line $LK$ is the radical axis of these two circles and every point on it including $T$ has equal powers, which is enough to finish.

Well known solution but $T$ is just the radical centre of point circles $A$;$M$ and $(AED)$.