Let ABC be any triangle and let D, E and F be the midpoints of AB, BC and CA. Let X be the point on BC such...
$begingroup$
Let ABC be any triangle and let D, E and F be the midpoints of AB, BC and CA. Let X be the point on BC such that AX is perpendicular to BC. Prove that X lies on the circumcircle of DEF.
Is it related to 9 points circle?
geometry
asked Jan 20 at 0:26
John AburiJohn Aburi
142
$endgroup$
add a comment |
$begingroup$
Let ABC be any triangle and let D, E and F be the midpoints of AB, BC and CA. Let X be the point on BC such that AX is perpendicular to BC. Prove that X lies on the circumcircle of DEF.
Is it related to 9 points circle?
geometry
asked Jan 20 at 0:26
John AburiJohn Aburi
142
$endgroup$
add a comment |
$begingroup$
Let ABC be any triangle and let D, E and F be the midpoints of AB, BC and CA. Let X be the point on BC such that AX is perpendicular to BC. Prove that X lies on the circumcircle of DEF.
Is it related to 9 points circle?
geometry
asked Jan 20 at 0:26
John AburiJohn Aburi
142
$endgroup$
Let ABC be any triangle and let D, E and F be the midpoints of AB, BC and CA. Let X be the point on BC such that AX is perpendicular to BC. Prove that X lies on the circumcircle of DEF.
Is it related to 9 points circle?
geometry
geometry
asked Jan 20 at 0:26
John AburiJohn Aburi
142
asked Jan 20 at 0:26
John AburiJohn Aburi
142
asked Jan 20 at 0:26
John AburiJohn Aburi
142
asked Jan 20 at 0:26
John AburiJohn Aburi
142
asked Jan 20 at 0:26
John AburiJohn Aburi
142
142
add a comment |
add a comment |
1 Answer
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$begingroup$
Consider the figure . All the $4$ small triangles are congruent , and all corresponding line segments are parallel . ( By Midpoint Theorem )
We have $angle FDE = angle C $ . Also , $angle FAX = 90 - angle C$
Since $FD parallel BC $ , $FG perp AX$ and bisects it .
$therefore triangle FAX $ is isosceles. It follows that $angle FXA = 90 - angle C$ .
Therefore , we have:- $$angle FXE + angle FDE = ( 90 + 90 - angle C) + angle C = 180 $$
Therefore , the opposite angles of quadrilateral $FXED$ are supplementary.
It follows that all the vertices lie on a circle .
Other cases , where the triangle is right or obtuse , can be proven by similar angle chasing .
Note:- Through this proof , we have proved that the points of intersection of the altitudes with the sides , and the midpoints of the sides , are all concyclic . We have proved that $6$ of the $9$ points of the $9$-point circle are concyclic. To prove the existence of the $9$-point circle , it remains to be shown that the midpoints of the lines joining the orthrocentre and the vertices are concyclic as well.
answered Jan 20 at 7:10
RahuboyRahuboy
63011
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider the figure . All the $4$ small triangles are congruent , and all corresponding line segments are parallel . ( By Midpoint Theorem )
We have $angle FDE = angle C $ . Also , $angle FAX = 90 - angle C$
Since $FD parallel BC $ , $FG perp AX$ and bisects it .
$therefore triangle FAX $ is isosceles. It follows that $angle FXA = 90 - angle C$ .
Therefore , we have:- $$angle FXE + angle FDE = ( 90 + 90 - angle C) + angle C = 180 $$
Therefore , the opposite angles of quadrilateral $FXED$ are supplementary.
It follows that all the vertices lie on a circle .
Other cases , where the triangle is right or obtuse , can be proven by similar angle chasing .
Note:- Through this proof , we have proved that the points of intersection of the altitudes with the sides , and the midpoints of the sides , are all concyclic . We have proved that $6$ of the $9$ points of the $9$-point circle are concyclic. To prove the existence of the $9$-point circle , it remains to be shown that the midpoints of the lines joining the orthrocentre and the vertices are concyclic as well.
answered Jan 20 at 7:10
RahuboyRahuboy
63011
$endgroup$
add a comment |
$begingroup$
Consider the figure . All the $4$ small triangles are congruent , and all corresponding line segments are parallel . ( By Midpoint Theorem )
We have $angle FDE = angle C $ . Also , $angle FAX = 90 - angle C$
Since $FD parallel BC $ , $FG perp AX$ and bisects it .
$therefore triangle FAX $ is isosceles. It follows that $angle FXA = 90 - angle C$ .
Therefore , we have:- $$angle FXE + angle FDE = ( 90 + 90 - angle C) + angle C = 180 $$
Therefore , the opposite angles of quadrilateral $FXED$ are supplementary.
It follows that all the vertices lie on a circle .
Other cases , where the triangle is right or obtuse , can be proven by similar angle chasing .
Note:- Through this proof , we have proved that the points of intersection of the altitudes with the sides , and the midpoints of the sides , are all concyclic . We have proved that $6$ of the $9$ points of the $9$-point circle are concyclic. To prove the existence of the $9$-point circle , it remains to be shown that the midpoints of the lines joining the orthrocentre and the vertices are concyclic as well.
answered Jan 20 at 7:10
RahuboyRahuboy
63011
$endgroup$
add a comment |
$begingroup$
Consider the figure . All the $4$ small triangles are congruent , and all corresponding line segments are parallel . ( By Midpoint Theorem )
We have $angle FDE = angle C $ . Also , $angle FAX = 90 - angle C$
Since $FD parallel BC $ , $FG perp AX$ and bisects it .
$therefore triangle FAX $ is isosceles. It follows that $angle FXA = 90 - angle C$ .
Therefore , we have:- $$angle FXE + angle FDE = ( 90 + 90 - angle C) + angle C = 180 $$
Therefore , the opposite angles of quadrilateral $FXED$ are supplementary.
It follows that all the vertices lie on a circle .
Other cases , where the triangle is right or obtuse , can be proven by similar angle chasing .
Note:- Through this proof , we have proved that the points of intersection of the altitudes with the sides , and the midpoints of the sides , are all concyclic . We have proved that $6$ of the $9$ points of the $9$-point circle are concyclic. To prove the existence of the $9$-point circle , it remains to be shown that the midpoints of the lines joining the orthrocentre and the vertices are concyclic as well.
answered Jan 20 at 7:10
RahuboyRahuboy
63011
$endgroup$
Consider the figure . All the $4$ small triangles are congruent , and all corresponding line segments are parallel . ( By Midpoint Theorem )
We have $angle FDE = angle C $ . Also , $angle FAX = 90 - angle C$
Since $FD parallel BC $ , $FG perp AX$ and bisects it .
$therefore triangle FAX $ is isosceles. It follows that $angle FXA = 90 - angle C$ .
Therefore , we have:- $$angle FXE + angle FDE = ( 90 + 90 - angle C) + angle C = 180 $$
Therefore , the opposite angles of quadrilateral $FXED$ are supplementary.
It follows that all the vertices lie on a circle .
Other cases , where the triangle is right or obtuse , can be proven by similar angle chasing .
Note:- Through this proof , we have proved that the points of intersection of the altitudes with the sides , and the midpoints of the sides , are all concyclic . We have proved that $6$ of the $9$ points of the $9$-point circle are concyclic. To prove the existence of the $9$-point circle , it remains to be shown that the midpoints of the lines joining the orthrocentre and the vertices are concyclic as well.
answered Jan 20 at 7:10
RahuboyRahuboy
63011
edited Jan 20 at 7:36
answered Jan 20 at 7:10
RahuboyRahuboy
63011
answered Jan 20 at 7:10
RahuboyRahuboy
63011
answered Jan 20 at 7:10
RahuboyRahuboy
63011
63011
add a comment |
add a comment |
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