Show that $N$ is orthcenter of triangle $AYZ$












4












$begingroup$


Triangle $ABC$ has height $BE,CF$.$M$ is the midpoint of $BC$. $EF$ and AM intersects at point $N$. Draw $NXperp BC, XYperp AB, XZperp AC(Xin BC, Yin AB, Zin AC)$.Show that $N$ is orthcenter of triangle $AYZ$.



I can't find any lemma to solve this problem. How we draw any another geometry element to solve.Show me pls! enter image description here






geometry







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$endgroup$








  • 2




    $begingroup$
    Can you attach a picture?
    $endgroup$
    – Love Invariants
    Dec 8 '18 at 12:34










  • $begingroup$
    I have added the picture. Pls show me this problem
    $endgroup$
    – Trong Tuan
    Dec 9 '18 at 1:47










  • $begingroup$
    What is the source of this problem?
    $endgroup$
    – Carl Schildkraut
    Dec 9 '18 at 3:09










  • $begingroup$
    Help me pls. I don't know where it is
    $endgroup$
    – Trong Tuan
    Dec 9 '18 at 3:42
















4












$begingroup$


Triangle $ABC$ has height $BE,CF$.$M$ is the midpoint of $BC$. $EF$ and AM intersects at point $N$. Draw $NXperp BC, XYperp AB, XZperp AC(Xin BC, Yin AB, Zin AC)$.Show that $N$ is orthcenter of triangle $AYZ$.



I can't find any lemma to solve this problem. How we draw any another geometry element to solve.Show me pls! enter image description here






geometry







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Can you attach a picture?
    $endgroup$
    – Love Invariants
    Dec 8 '18 at 12:34










  • $begingroup$
    I have added the picture. Pls show me this problem
    $endgroup$
    – Trong Tuan
    Dec 9 '18 at 1:47










  • $begingroup$
    What is the source of this problem?
    $endgroup$
    – Carl Schildkraut
    Dec 9 '18 at 3:09










  • $begingroup$
    Help me pls. I don't know where it is
    $endgroup$
    – Trong Tuan
    Dec 9 '18 at 3:42














4












4








4


3



$begingroup$


Triangle $ABC$ has height $BE,CF$.$M$ is the midpoint of $BC$. $EF$ and AM intersects at point $N$. Draw $NXperp BC, XYperp AB, XZperp AC(Xin BC, Yin AB, Zin AC)$.Show that $N$ is orthcenter of triangle $AYZ$.



I can't find any lemma to solve this problem. How we draw any another geometry element to solve.Show me pls! enter image description here






geometry







share|cite|improve this question











$endgroup$




Triangle $ABC$ has height $BE,CF$.$M$ is the midpoint of $BC$. $EF$ and AM intersects at point $N$. Draw $NXperp BC, XYperp AB, XZperp AC(Xin BC, Yin AB, Zin AC)$.Show that $N$ is orthcenter of triangle $AYZ$.



I can't find any lemma to solve this problem. How we draw any another geometry element to solve.Show me pls! enter image description here






geometry




geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 1:49







Trong Tuan

















asked Dec 8 '18 at 12:04









Trong TuanTrong Tuan

1268




1268








  • 2




    $begingroup$
    Can you attach a picture?
    $endgroup$
    – Love Invariants
    Dec 8 '18 at 12:34










  • $begingroup$
    I have added the picture. Pls show me this problem
    $endgroup$
    – Trong Tuan
    Dec 9 '18 at 1:47










  • $begingroup$
    What is the source of this problem?
    $endgroup$
    – Carl Schildkraut
    Dec 9 '18 at 3:09










  • $begingroup$
    Help me pls. I don't know where it is
    $endgroup$
    – Trong Tuan
    Dec 9 '18 at 3:42














  • 2




    $begingroup$
    Can you attach a picture?
    $endgroup$
    – Love Invariants
    Dec 8 '18 at 12:34










  • $begingroup$
    I have added the picture. Pls show me this problem
    $endgroup$
    – Trong Tuan
    Dec 9 '18 at 1:47










  • $begingroup$
    What is the source of this problem?
    $endgroup$
    – Carl Schildkraut
    Dec 9 '18 at 3:09










  • $begingroup$
    Help me pls. I don't know where it is
    $endgroup$
    – Trong Tuan
    Dec 9 '18 at 3:42








2




2




$begingroup$
Can you attach a picture?
$endgroup$
– Love Invariants
Dec 8 '18 at 12:34




$begingroup$
Can you attach a picture?
$endgroup$
– Love Invariants
Dec 8 '18 at 12:34












$begingroup$
I have added the picture. Pls show me this problem
$endgroup$
– Trong Tuan
Dec 9 '18 at 1:47




$begingroup$
I have added the picture. Pls show me this problem
$endgroup$
– Trong Tuan
Dec 9 '18 at 1:47












$begingroup$
What is the source of this problem?
$endgroup$
– Carl Schildkraut
Dec 9 '18 at 3:09




$begingroup$
What is the source of this problem?
$endgroup$
– Carl Schildkraut
Dec 9 '18 at 3:09












$begingroup$
Help me pls. I don't know where it is
$endgroup$
– Trong Tuan
Dec 9 '18 at 3:42




$begingroup$
Help me pls. I don't know where it is
$endgroup$
– Trong Tuan
Dec 9 '18 at 3:42










1 Answer
1






active

oldest

votes


















2












$begingroup$

enter image description here
First note that $triangle AEF$ and $triangle ABC$ are similar. Let $P$ be the midpoint of $EF$, then $$angle APE = angle AMBtag{1}.$$
Moreover, $E,F$ lie on the circle with center $M$ and diameter $BC$, so $MPperp EF$. So $MNPX$ is circular. It follows that
$$angle NPX + angle AMB = 180^circ.tag{2}$$
From (1) and (2) we conclude that $A,P,X$ are colinear. Consequently,
$$frac{BX}{XC} = frac{NE}{NF}.tag{3}$$
Since $XZparallel BE$, we have $EZ : ZC = BX:XC = NE:NF$. Thus $NZparallel CF$, and similarly $NY parallel BE$ and we are done.




Note: (3) may be a little subtle but I leave it to you to clarify.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your solution is nice. But it lacks proper explanation. Might be not so good for newbies. You jumped to so many conclusions.
    $endgroup$
    – Love Invariants
    Dec 9 '18 at 18:14











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1 Answer
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1 Answer
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2












$begingroup$

enter image description here
First note that $triangle AEF$ and $triangle ABC$ are similar. Let $P$ be the midpoint of $EF$, then $$angle APE = angle AMBtag{1}.$$
Moreover, $E,F$ lie on the circle with center $M$ and diameter $BC$, so $MPperp EF$. So $MNPX$ is circular. It follows that
$$angle NPX + angle AMB = 180^circ.tag{2}$$
From (1) and (2) we conclude that $A,P,X$ are colinear. Consequently,
$$frac{BX}{XC} = frac{NE}{NF}.tag{3}$$
Since $XZparallel BE$, we have $EZ : ZC = BX:XC = NE:NF$. Thus $NZparallel CF$, and similarly $NY parallel BE$ and we are done.




Note: (3) may be a little subtle but I leave it to you to clarify.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your solution is nice. But it lacks proper explanation. Might be not so good for newbies. You jumped to so many conclusions.
    $endgroup$
    – Love Invariants
    Dec 9 '18 at 18:14
















2












$begingroup$

enter image description here
First note that $triangle AEF$ and $triangle ABC$ are similar. Let $P$ be the midpoint of $EF$, then $$angle APE = angle AMBtag{1}.$$
Moreover, $E,F$ lie on the circle with center $M$ and diameter $BC$, so $MPperp EF$. So $MNPX$ is circular. It follows that
$$angle NPX + angle AMB = 180^circ.tag{2}$$
From (1) and (2) we conclude that $A,P,X$ are colinear. Consequently,
$$frac{BX}{XC} = frac{NE}{NF}.tag{3}$$
Since $XZparallel BE$, we have $EZ : ZC = BX:XC = NE:NF$. Thus $NZparallel CF$, and similarly $NY parallel BE$ and we are done.




Note: (3) may be a little subtle but I leave it to you to clarify.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your solution is nice. But it lacks proper explanation. Might be not so good for newbies. You jumped to so many conclusions.
    $endgroup$
    – Love Invariants
    Dec 9 '18 at 18:14














2












2








2





$begingroup$

enter image description here
First note that $triangle AEF$ and $triangle ABC$ are similar. Let $P$ be the midpoint of $EF$, then $$angle APE = angle AMBtag{1}.$$
Moreover, $E,F$ lie on the circle with center $M$ and diameter $BC$, so $MPperp EF$. So $MNPX$ is circular. It follows that
$$angle NPX + angle AMB = 180^circ.tag{2}$$
From (1) and (2) we conclude that $A,P,X$ are colinear. Consequently,
$$frac{BX}{XC} = frac{NE}{NF}.tag{3}$$
Since $XZparallel BE$, we have $EZ : ZC = BX:XC = NE:NF$. Thus $NZparallel CF$, and similarly $NY parallel BE$ and we are done.




Note: (3) may be a little subtle but I leave it to you to clarify.






share|cite|improve this answer









$endgroup$



enter image description here
First note that $triangle AEF$ and $triangle ABC$ are similar. Let $P$ be the midpoint of $EF$, then $$angle APE = angle AMBtag{1}.$$
Moreover, $E,F$ lie on the circle with center $M$ and diameter $BC$, so $MPperp EF$. So $MNPX$ is circular. It follows that
$$angle NPX + angle AMB = 180^circ.tag{2}$$
From (1) and (2) we conclude that $A,P,X$ are colinear. Consequently,
$$frac{BX}{XC} = frac{NE}{NF}.tag{3}$$
Since $XZparallel BE$, we have $EZ : ZC = BX:XC = NE:NF$. Thus $NZparallel CF$, and similarly $NY parallel BE$ and we are done.




Note: (3) may be a little subtle but I leave it to you to clarify.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 4:16









Quang HoangQuang Hoang

12.8k1132




12.8k1132












  • $begingroup$
    Your solution is nice. But it lacks proper explanation. Might be not so good for newbies. You jumped to so many conclusions.
    $endgroup$
    – Love Invariants
    Dec 9 '18 at 18:14


















  • $begingroup$
    Your solution is nice. But it lacks proper explanation. Might be not so good for newbies. You jumped to so many conclusions.
    $endgroup$
    – Love Invariants
    Dec 9 '18 at 18:14
















$begingroup$
Your solution is nice. But it lacks proper explanation. Might be not so good for newbies. You jumped to so many conclusions.
$endgroup$
– Love Invariants
Dec 9 '18 at 18:14




$begingroup$
Your solution is nice. But it lacks proper explanation. Might be not so good for newbies. You jumped to so many conclusions.
$endgroup$
– Love Invariants
Dec 9 '18 at 18:14


















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