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Prove MB bisects angle IMF
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$Delta ABC$ inscribed $(O)$, incenter $I$, $(I)$ touch $BC$ at $D$. $AD cap (O)=$$A;E$. Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of arc $BAC$. Let $EN$ intersect $(BIC)$ at $G$ ($G$ lies in $ABC$). $(AGE) cap (BIC)=$$G;F$. Prove $MB$ bisects $angle IMF$
I have proved $overlineG,D,F$ by using power of a point. I need to prove $I,D,G,M$ concyclic (haven't proved yet), then the problem will be much easier ... Any ideas ?
euclidean-geometry
asked Aug 1 at 6:52
RopuToran
976
add a comment |Â
up vote
2
down vote
favorite
$Delta ABC$ inscribed $(O)$, incenter $I$, $(I)$ touch $BC$ at $D$. $AD cap (O)=$$A;E$. Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of arc $BAC$. Let $EN$ intersect $(BIC)$ at $G$ ($G$ lies in $ABC$). $(AGE) cap (BIC)=$$G;F$. Prove $MB$ bisects $angle IMF$
I have proved $overlineG,D,F$ by using power of a point. I need to prove $I,D,G,M$ concyclic (haven't proved yet), then the problem will be much easier ... Any ideas ?
euclidean-geometry
asked Aug 1 at 6:52
RopuToran
976
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$Delta ABC$ inscribed $(O)$, incenter $I$, $(I)$ touch $BC$ at $D$. $AD cap (O)=$$A;E$. Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of arc $BAC$. Let $EN$ intersect $(BIC)$ at $G$ ($G$ lies in $ABC$). $(AGE) cap (BIC)=$$G;F$. Prove $MB$ bisects $angle IMF$
I have proved $overlineG,D,F$ by using power of a point. I need to prove $I,D,G,M$ concyclic (haven't proved yet), then the problem will be much easier ... Any ideas ?
euclidean-geometry
asked Aug 1 at 6:52
RopuToran
976
$Delta ABC$ inscribed $(O)$, incenter $I$, $(I)$ touch $BC$ at $D$. $AD cap (O)=$$A;E$. Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of arc $BAC$. Let $EN$ intersect $(BIC)$ at $G$ ($G$ lies in $ABC$). $(AGE) cap (BIC)=$$G;F$. Prove $MB$ bisects $angle IMF$
I have proved $overlineG,D,F$ by using power of a point. I need to prove $I,D,G,M$ concyclic (haven't proved yet), then the problem will be much easier ... Any ideas ?
euclidean-geometry
asked Aug 1 at 6:52
RopuToran
976
asked Aug 1 at 6:52
RopuToran
976
asked Aug 1 at 6:52
RopuToran
976
asked Aug 1 at 6:52
RopuToran
976
976
add a comment |Â
add a comment |Â
1 Answer
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Since $angle IDM = 90^0$, IM is the diameter of (IDM). Let L be a point somewhere near F. LI can be drawn tangent to (IDM) at I. Then, $angle IMD = angle LID$.
Produce IM to K such that IM = MK. Then, M is the center of the circle passing I, and K cutting MF at H. Note that LI is also tangent to circle (IHK) at I. Then, $angle LIH = angle IKH$.
Combining the two results, we can say that (1) BDMC // HK and (2) I, D, H are collinear. By midpoint theorem, ID = DH. Required result follows.
answered Aug 1 at 15:53
Mick
11.5k21540
I assume you meant $angle LIH=angle IHK$. However, there does seem to be circular logic somewhere. To know that $B$ and $HK$ are parallel, you need to know that $I$, $D$, and $H$ are collinear. And you don't seem to be using how the points are constructed,.
– Batominovski
Aug 1 at 16:02
@Batominovski No, it is $angle LIH = angle IKH$. Perhaps an added picture can make things more clear. I, D, and H are collinear because of (1); plus $angle IDM = angle IHK = 90^0$; and the two angles are in corresponding positions.
– Mick
Aug 1 at 18:52
What do you mean "Let L be a point somewhere near F" ?
– RopuToran
Aug 2 at 17:46
@RopuToran I just want to draw a line that is tangent to the circle IDM touching it at I. That is the blue line. I need to give that tangent line a name and therefore LI. L is also used (in the later stage) for naming some angles. For example, $angle LID$.
– Mick
Aug 2 at 18:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Since $angle IDM = 90^0$, IM is the diameter of (IDM). Let L be a point somewhere near F. LI can be drawn tangent to (IDM) at I. Then, $angle IMD = angle LID$.
Produce IM to K such that IM = MK. Then, M is the center of the circle passing I, and K cutting MF at H. Note that LI is also tangent to circle (IHK) at I. Then, $angle LIH = angle IKH$.
Combining the two results, we can say that (1) BDMC // HK and (2) I, D, H are collinear. By midpoint theorem, ID = DH. Required result follows.
answered Aug 1 at 15:53
Mick
11.5k21540
I assume you meant $angle LIH=angle IHK$. However, there does seem to be circular logic somewhere. To know that $B$ and $HK$ are parallel, you need to know that $I$, $D$, and $H$ are collinear. And you don't seem to be using how the points are constructed,.
– Batominovski
Aug 1 at 16:02
@Batominovski No, it is $angle LIH = angle IKH$. Perhaps an added picture can make things more clear. I, D, and H are collinear because of (1); plus $angle IDM = angle IHK = 90^0$; and the two angles are in corresponding positions.
– Mick
Aug 1 at 18:52
What do you mean "Let L be a point somewhere near F" ?
– RopuToran
Aug 2 at 17:46
@RopuToran I just want to draw a line that is tangent to the circle IDM touching it at I. That is the blue line. I need to give that tangent line a name and therefore LI. L is also used (in the later stage) for naming some angles. For example, $angle LID$.
– Mick
Aug 2 at 18:04
add a comment |Â
up vote
0
down vote
Since $angle IDM = 90^0$, IM is the diameter of (IDM). Let L be a point somewhere near F. LI can be drawn tangent to (IDM) at I. Then, $angle IMD = angle LID$.
Produce IM to K such that IM = MK. Then, M is the center of the circle passing I, and K cutting MF at H. Note that LI is also tangent to circle (IHK) at I. Then, $angle LIH = angle IKH$.
Combining the two results, we can say that (1) BDMC // HK and (2) I, D, H are collinear. By midpoint theorem, ID = DH. Required result follows.
answered Aug 1 at 15:53
Mick
11.5k21540
I assume you meant $angle LIH=angle IHK$. However, there does seem to be circular logic somewhere. To know that $B$ and $HK$ are parallel, you need to know that $I$, $D$, and $H$ are collinear. And you don't seem to be using how the points are constructed,.
– Batominovski
Aug 1 at 16:02
@Batominovski No, it is $angle LIH = angle IKH$. Perhaps an added picture can make things more clear. I, D, and H are collinear because of (1); plus $angle IDM = angle IHK = 90^0$; and the two angles are in corresponding positions.
– Mick
Aug 1 at 18:52
What do you mean "Let L be a point somewhere near F" ?
– RopuToran
Aug 2 at 17:46
@RopuToran I just want to draw a line that is tangent to the circle IDM touching it at I. That is the blue line. I need to give that tangent line a name and therefore LI. L is also used (in the later stage) for naming some angles. For example, $angle LID$.
– Mick
Aug 2 at 18:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $angle IDM = 90^0$, IM is the diameter of (IDM). Let L be a point somewhere near F. LI can be drawn tangent to (IDM) at I. Then, $angle IMD = angle LID$.
Produce IM to K such that IM = MK. Then, M is the center of the circle passing I, and K cutting MF at H. Note that LI is also tangent to circle (IHK) at I. Then, $angle LIH = angle IKH$.
Combining the two results, we can say that (1) BDMC // HK and (2) I, D, H are collinear. By midpoint theorem, ID = DH. Required result follows.
answered Aug 1 at 15:53
Mick
11.5k21540
Since $angle IDM = 90^0$, IM is the diameter of (IDM). Let L be a point somewhere near F. LI can be drawn tangent to (IDM) at I. Then, $angle IMD = angle LID$.
Produce IM to K such that IM = MK. Then, M is the center of the circle passing I, and K cutting MF at H. Note that LI is also tangent to circle (IHK) at I. Then, $angle LIH = angle IKH$.
Combining the two results, we can say that (1) BDMC // HK and (2) I, D, H are collinear. By midpoint theorem, ID = DH. Required result follows.
answered Aug 1 at 15:53
Mick
11.5k21540
edited Aug 1 at 18:56
answered Aug 1 at 15:53
Mick
11.5k21540
answered Aug 1 at 15:53
Mick
11.5k21540
answered Aug 1 at 15:53
Mick
11.5k21540
11.5k21540
I assume you meant $angle LIH=angle IHK$. However, there does seem to be circular logic somewhere. To know that $B$ and $HK$ are parallel, you need to know that $I$, $D$, and $H$ are collinear. And you don't seem to be using how the points are constructed,.
– Batominovski
Aug 1 at 16:02
@Batominovski No, it is $angle LIH = angle IKH$. Perhaps an added picture can make things more clear. I, D, and H are collinear because of (1); plus $angle IDM = angle IHK = 90^0$; and the two angles are in corresponding positions.
– Mick
Aug 1 at 18:52
What do you mean "Let L be a point somewhere near F" ?
– RopuToran
Aug 2 at 17:46
@RopuToran I just want to draw a line that is tangent to the circle IDM touching it at I. That is the blue line. I need to give that tangent line a name and therefore LI. L is also used (in the later stage) for naming some angles. For example, $angle LID$.
– Mick
Aug 2 at 18:04
add a comment |Â
I assume you meant $angle LIH=angle IHK$. However, there does seem to be circular logic somewhere. To know that $B$ and $HK$ are parallel, you need to know that $I$, $D$, and $H$ are collinear. And you don't seem to be using how the points are constructed,.
– Batominovski
Aug 1 at 16:02
@Batominovski No, it is $angle LIH = angle IKH$. Perhaps an added picture can make things more clear. I, D, and H are collinear because of (1); plus $angle IDM = angle IHK = 90^0$; and the two angles are in corresponding positions.
– Mick
Aug 1 at 18:52
What do you mean "Let L be a point somewhere near F" ?
– RopuToran
Aug 2 at 17:46
@RopuToran I just want to draw a line that is tangent to the circle IDM touching it at I. That is the blue line. I need to give that tangent line a name and therefore LI. L is also used (in the later stage) for naming some angles. For example, $angle LID$.
– Mick
Aug 2 at 18:04
I assume you meant $angle LIH=angle IHK$. However, there does seem to be circular logic somewhere. To know that $B$ and $HK$ are parallel, you need to know that $I$, $D$, and $H$ are collinear. And you don't seem to be using how the points are constructed,.
– Batominovski
Aug 1 at 16:02
I assume you meant $angle LIH=angle IHK$. However, there does seem to be circular logic somewhere. To know that $B$ and $HK$ are parallel, you need to know that $I$, $D$, and $H$ are collinear. And you don't seem to be using how the points are constructed,.
– Batominovski
Aug 1 at 16:02
@Batominovski No, it is $angle LIH = angle IKH$. Perhaps an added picture can make things more clear. I, D, and H are collinear because of (1); plus $angle IDM = angle IHK = 90^0$; and the two angles are in corresponding positions.
– Mick
Aug 1 at 18:52
@Batominovski No, it is $angle LIH = angle IKH$. Perhaps an added picture can make things more clear. I, D, and H are collinear because of (1); plus $angle IDM = angle IHK = 90^0$; and the two angles are in corresponding positions.
– Mick
Aug 1 at 18:52
What do you mean "Let L be a point somewhere near F" ?
– RopuToran
Aug 2 at 17:46
What do you mean "Let L be a point somewhere near F" ?
– RopuToran
Aug 2 at 17:46
@RopuToran I just want to draw a line that is tangent to the circle IDM touching it at I. That is the blue line. I need to give that tangent line a name and therefore LI. L is also used (in the later stage) for naming some angles. For example, $angle LID$.
– Mick
Aug 2 at 18:04
@RopuToran I just want to draw a line that is tangent to the circle IDM touching it at I. That is the blue line. I need to give that tangent line a name and therefore LI. L is also used (in the later stage) for naming some angles. For example, $angle LID$.
– Mick
Aug 2 at 18:04
add a comment |Â
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