Prove that $CE=AB$
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Suppose $ABC$ is an acute-angled triangle with $AB<AC$.Let $M$ be the midpoint of $BC$. Suppose $P$ is a point on side $AB$ such that, if $PC$ intersects the median $AM$ at E, then $AP=PE$.Prove that $AB=CE$.
I don't know how to start. Please give me an idea. Getting no fruitful thoughts ,I started using barycentric Coordinates.But the calculations seemed very tough and I failed to proceed. Please give me any idea to start
geometry euclidean-geometry triangle
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up vote
2
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Suppose $ABC$ is an acute-angled triangle with $AB<AC$.Let $M$ be the midpoint of $BC$. Suppose $P$ is a point on side $AB$ such that, if $PC$ intersects the median $AM$ at E, then $AP=PE$.Prove that $AB=CE$.
I don't know how to start. Please give me an idea. Getting no fruitful thoughts ,I started using barycentric Coordinates.But the calculations seemed very tough and I failed to proceed. Please give me any idea to start
geometry euclidean-geometry triangle
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $ABC$ is an acute-angled triangle with $AB<AC$.Let $M$ be the midpoint of $BC$. Suppose $P$ is a point on side $AB$ such that, if $PC$ intersects the median $AM$ at E, then $AP=PE$.Prove that $AB=CE$.
I don't know how to start. Please give me an idea. Getting no fruitful thoughts ,I started using barycentric Coordinates.But the calculations seemed very tough and I failed to proceed. Please give me any idea to start
geometry euclidean-geometry triangle
Suppose $ABC$ is an acute-angled triangle with $AB<AC$.Let $M$ be the midpoint of $BC$. Suppose $P$ is a point on side $AB$ such that, if $PC$ intersects the median $AM$ at E, then $AP=PE$.Prove that $AB=CE$.
I don't know how to start. Please give me an idea. Getting no fruitful thoughts ,I started using barycentric Coordinates.But the calculations seemed very tough and I failed to proceed. Please give me any idea to start
geometry euclidean-geometry triangle
asked Jul 23 at 17:23
Sufaid Saleel
1,666625
1,666625
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2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Hint: Â write Menelaus' theorem for triangle $,triangle PBC,$ and transversal $,AM,$.
add a comment |Â
up vote
2
down vote
Construct the //gm BECF.
From (1) AEM is a straight line; (2) E is a vertex of that //gm; (3) M is the midpoint of one of its diagonals; and (4) F is a vertex of that //gm and FM is a straight line, we can say that AEMF is a straight line.
The required result follows because all the green marked angles are equal.
What does //gm mean?
– greedoid
Jul 23 at 19:02
@Angle I guess it stands for Parallelogram?
– Mythomorphic
Jul 23 at 19:13
Nice solution +1
– greedoid
Jul 23 at 19:25
@Mythomorphic That is right.
– Mick
Jul 24 at 2:45
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint: Â write Menelaus' theorem for triangle $,triangle PBC,$ and transversal $,AM,$.
add a comment |Â
up vote
4
down vote
accepted
Hint: Â write Menelaus' theorem for triangle $,triangle PBC,$ and transversal $,AM,$.
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint: Â write Menelaus' theorem for triangle $,triangle PBC,$ and transversal $,AM,$.
Hint: Â write Menelaus' theorem for triangle $,triangle PBC,$ and transversal $,AM,$.
answered Jul 23 at 17:31


dxiv
54k64796
54k64796
add a comment |Â
add a comment |Â
up vote
2
down vote
Construct the //gm BECF.
From (1) AEM is a straight line; (2) E is a vertex of that //gm; (3) M is the midpoint of one of its diagonals; and (4) F is a vertex of that //gm and FM is a straight line, we can say that AEMF is a straight line.
The required result follows because all the green marked angles are equal.
What does //gm mean?
– greedoid
Jul 23 at 19:02
@Angle I guess it stands for Parallelogram?
– Mythomorphic
Jul 23 at 19:13
Nice solution +1
– greedoid
Jul 23 at 19:25
@Mythomorphic That is right.
– Mick
Jul 24 at 2:45
add a comment |Â
up vote
2
down vote
Construct the //gm BECF.
From (1) AEM is a straight line; (2) E is a vertex of that //gm; (3) M is the midpoint of one of its diagonals; and (4) F is a vertex of that //gm and FM is a straight line, we can say that AEMF is a straight line.
The required result follows because all the green marked angles are equal.
What does //gm mean?
– greedoid
Jul 23 at 19:02
@Angle I guess it stands for Parallelogram?
– Mythomorphic
Jul 23 at 19:13
Nice solution +1
– greedoid
Jul 23 at 19:25
@Mythomorphic That is right.
– Mick
Jul 24 at 2:45
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Construct the //gm BECF.
From (1) AEM is a straight line; (2) E is a vertex of that //gm; (3) M is the midpoint of one of its diagonals; and (4) F is a vertex of that //gm and FM is a straight line, we can say that AEMF is a straight line.
The required result follows because all the green marked angles are equal.
Construct the //gm BECF.
From (1) AEM is a straight line; (2) E is a vertex of that //gm; (3) M is the midpoint of one of its diagonals; and (4) F is a vertex of that //gm and FM is a straight line, we can say that AEMF is a straight line.
The required result follows because all the green marked angles are equal.
edited Jul 23 at 18:53
answered Jul 23 at 18:45


Mick
11.5k21540
11.5k21540
What does //gm mean?
– greedoid
Jul 23 at 19:02
@Angle I guess it stands for Parallelogram?
– Mythomorphic
Jul 23 at 19:13
Nice solution +1
– greedoid
Jul 23 at 19:25
@Mythomorphic That is right.
– Mick
Jul 24 at 2:45
add a comment |Â
What does //gm mean?
– greedoid
Jul 23 at 19:02
@Angle I guess it stands for Parallelogram?
– Mythomorphic
Jul 23 at 19:13
Nice solution +1
– greedoid
Jul 23 at 19:25
@Mythomorphic That is right.
– Mick
Jul 24 at 2:45
What does //gm mean?
– greedoid
Jul 23 at 19:02
What does //gm mean?
– greedoid
Jul 23 at 19:02
@Angle I guess it stands for Parallelogram?
– Mythomorphic
Jul 23 at 19:13
@Angle I guess it stands for Parallelogram?
– Mythomorphic
Jul 23 at 19:13
Nice solution +1
– greedoid
Jul 23 at 19:25
Nice solution +1
– greedoid
Jul 23 at 19:25
@Mythomorphic That is right.
– Mick
Jul 24 at 2:45
@Mythomorphic That is right.
– Mick
Jul 24 at 2:45
add a comment |Â
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