High school geometry problem: Reflect a vertex about opposite side.
$begingroup$
Let $ABC$ be a triangle and $A'$ be the reflection of $A$ about $BC$. Let $M$ be the mid-point of $BC$ and $I$ be the point of intersection of $A'M$ with the circumcircle of triangle $ABC$ (See figure).
To prove that: $angle ABC=angle MIB$.
I am not able to make any progress.
geometry contest-math euclidean-geometry reflection geometric-transformation
edited Jan 8 at 18:41
greedoid
42.9k1153105
asked Dec 27 '18 at 14:49
caffeinemachinecaffeinemachine
6,57221351
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be a triangle and $A'$ be the reflection of $A$ about $BC$. Let $M$ be the mid-point of $BC$ and $I$ be the point of intersection of $A'M$ with the circumcircle of triangle $ABC$ (See figure).
To prove that: $angle ABC=angle MIB$.
I am not able to make any progress.
geometry contest-math euclidean-geometry reflection geometric-transformation
edited Jan 8 at 18:41
greedoid
42.9k1153105
asked Dec 27 '18 at 14:49
caffeinemachinecaffeinemachine
6,57221351
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be a triangle and $A'$ be the reflection of $A$ about $BC$. Let $M$ be the mid-point of $BC$ and $I$ be the point of intersection of $A'M$ with the circumcircle of triangle $ABC$ (See figure).
To prove that: $angle ABC=angle MIB$.
I am not able to make any progress.
geometry contest-math euclidean-geometry reflection geometric-transformation
edited Jan 8 at 18:41
greedoid
42.9k1153105
asked Dec 27 '18 at 14:49
caffeinemachinecaffeinemachine
6,57221351
$endgroup$
Let $ABC$ be a triangle and $A'$ be the reflection of $A$ about $BC$. Let $M$ be the mid-point of $BC$ and $I$ be the point of intersection of $A'M$ with the circumcircle of triangle $ABC$ (See figure).
To prove that: $angle ABC=angle MIB$.
I am not able to make any progress.
geometry contest-math euclidean-geometry reflection geometric-transformation
geometry contest-math euclidean-geometry reflection geometric-transformation
edited Jan 8 at 18:41
greedoid
42.9k1153105
asked Dec 27 '18 at 14:49
caffeinemachinecaffeinemachine
6,57221351
edited Jan 8 at 18:41
greedoid
42.9k1153105
asked Dec 27 '18 at 14:49
caffeinemachinecaffeinemachine
6,57221351
edited Jan 8 at 18:41
greedoid
42.9k1153105
edited Jan 8 at 18:41
greedoid
42.9k1153105
edited Jan 8 at 18:41
greedoid
42.9k1153105
42.9k1153105
asked Dec 27 '18 at 14:49
caffeinemachinecaffeinemachine
6,57221351
asked Dec 27 '18 at 14:49
caffeinemachinecaffeinemachine
6,57221351
asked Dec 27 '18 at 14:49
caffeinemachinecaffeinemachine
6,57221351
6,57221351
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $AM$ cuts circle second time at $J$. Then by symmetry with respect to perpendicular bisector of $BC$, the line $AM$ goes to line $MA'$ and circle goes to it self (also $B$ to $C$ and vice versa, and $M$ to it self), so $I$ goes to $J$. Thus $$angle ABC = angle AJC = angle MJC = angle MIB$$
answered Dec 27 '18 at 15:24
greedoidgreedoid
42.9k1153105
$endgroup$
$begingroup$
Very nice proof.
$endgroup$
– caffeinemachine
Dec 27 '18 at 16:25
$begingroup$
we are asked to prove angle ABC = angle MIB not angle MIC
$endgroup$
– Lozenges
Dec 27 '18 at 16:33
$begingroup$
also how do we know MA goes to MA' ?
$endgroup$
– Lozenges
Dec 27 '18 at 16:41
$begingroup$
@Lozenges What greedoid means is that under the transformation of the plane "reflection about the perpendicular bisector of BC", the line $MA$ transforms to $MA'$.
$endgroup$
– caffeinemachine
Dec 27 '18 at 17:17
$begingroup$
@caffeinemachine How do we know this? M is fixed , A goes to some point other than A'. MA' is symmetric of MA with respect to BC. How is it symmetric with respect to the perpendicular bisector of BC?
$endgroup$
– Lozenges
Dec 27 '18 at 17:25
|
show 2 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $AM$ cuts circle second time at $J$. Then by symmetry with respect to perpendicular bisector of $BC$, the line $AM$ goes to line $MA'$ and circle goes to it self (also $B$ to $C$ and vice versa, and $M$ to it self), so $I$ goes to $J$. Thus $$angle ABC = angle AJC = angle MJC = angle MIB$$
answered Dec 27 '18 at 15:24
greedoidgreedoid
42.9k1153105
$endgroup$
$begingroup$
Very nice proof.
$endgroup$
– caffeinemachine
Dec 27 '18 at 16:25
$begingroup$
we are asked to prove angle ABC = angle MIB not angle MIC
$endgroup$
– Lozenges
Dec 27 '18 at 16:33
$begingroup$
also how do we know MA goes to MA' ?
$endgroup$
– Lozenges
Dec 27 '18 at 16:41
$begingroup$
@Lozenges What greedoid means is that under the transformation of the plane "reflection about the perpendicular bisector of BC", the line $MA$ transforms to $MA'$.
$endgroup$
– caffeinemachine
Dec 27 '18 at 17:17
$begingroup$
@caffeinemachine How do we know this? M is fixed , A goes to some point other than A'. MA' is symmetric of MA with respect to BC. How is it symmetric with respect to the perpendicular bisector of BC?
$endgroup$
– Lozenges
Dec 27 '18 at 17:25
|
show 2 more comments
$begingroup$
Let $AM$ cuts circle second time at $J$. Then by symmetry with respect to perpendicular bisector of $BC$, the line $AM$ goes to line $MA'$ and circle goes to it self (also $B$ to $C$ and vice versa, and $M$ to it self), so $I$ goes to $J$. Thus $$angle ABC = angle AJC = angle MJC = angle MIB$$
answered Dec 27 '18 at 15:24
greedoidgreedoid
42.9k1153105
$endgroup$
$begingroup$
Very nice proof.
$endgroup$
– caffeinemachine
Dec 27 '18 at 16:25
$begingroup$
we are asked to prove angle ABC = angle MIB not angle MIC
$endgroup$
– Lozenges
Dec 27 '18 at 16:33
$begingroup$
also how do we know MA goes to MA' ?
$endgroup$
– Lozenges
Dec 27 '18 at 16:41
$begingroup$
@Lozenges What greedoid means is that under the transformation of the plane "reflection about the perpendicular bisector of BC", the line $MA$ transforms to $MA'$.
$endgroup$
– caffeinemachine
Dec 27 '18 at 17:17
$begingroup$
@caffeinemachine How do we know this? M is fixed , A goes to some point other than A'. MA' is symmetric of MA with respect to BC. How is it symmetric with respect to the perpendicular bisector of BC?
$endgroup$
– Lozenges
Dec 27 '18 at 17:25
|
show 2 more comments
$begingroup$
Let $AM$ cuts circle second time at $J$. Then by symmetry with respect to perpendicular bisector of $BC$, the line $AM$ goes to line $MA'$ and circle goes to it self (also $B$ to $C$ and vice versa, and $M$ to it self), so $I$ goes to $J$. Thus $$angle ABC = angle AJC = angle MJC = angle MIB$$
answered Dec 27 '18 at 15:24
greedoidgreedoid
42.9k1153105
$endgroup$
Let $AM$ cuts circle second time at $J$. Then by symmetry with respect to perpendicular bisector of $BC$, the line $AM$ goes to line $MA'$ and circle goes to it self (also $B$ to $C$ and vice versa, and $M$ to it self), so $I$ goes to $J$. Thus $$angle ABC = angle AJC = angle MJC = angle MIB$$
answered Dec 27 '18 at 15:24
greedoidgreedoid
42.9k1153105
edited Dec 27 '18 at 16:42
answered Dec 27 '18 at 15:24
greedoidgreedoid
42.9k1153105
answered Dec 27 '18 at 15:24
greedoidgreedoid
42.9k1153105
answered Dec 27 '18 at 15:24
greedoidgreedoid
42.9k1153105
42.9k1153105
$begingroup$
Very nice proof.
$endgroup$
– caffeinemachine
Dec 27 '18 at 16:25
$begingroup$
we are asked to prove angle ABC = angle MIB not angle MIC
$endgroup$
– Lozenges
Dec 27 '18 at 16:33
$begingroup$
also how do we know MA goes to MA' ?
$endgroup$
– Lozenges
Dec 27 '18 at 16:41
$begingroup$
@Lozenges What greedoid means is that under the transformation of the plane "reflection about the perpendicular bisector of BC", the line $MA$ transforms to $MA'$.
$endgroup$
– caffeinemachine
Dec 27 '18 at 17:17
$begingroup$
@caffeinemachine How do we know this? M is fixed , A goes to some point other than A'. MA' is symmetric of MA with respect to BC. How is it symmetric with respect to the perpendicular bisector of BC?
$endgroup$
– Lozenges
Dec 27 '18 at 17:25
|
show 2 more comments
$begingroup$
Very nice proof.
$endgroup$
– caffeinemachine
Dec 27 '18 at 16:25
$begingroup$
we are asked to prove angle ABC = angle MIB not angle MIC
$endgroup$
– Lozenges
Dec 27 '18 at 16:33
$begingroup$
also how do we know MA goes to MA' ?
$endgroup$
– Lozenges
Dec 27 '18 at 16:41
$begingroup$
@Lozenges What greedoid means is that under the transformation of the plane "reflection about the perpendicular bisector of BC", the line $MA$ transforms to $MA'$.
$endgroup$
– caffeinemachine
Dec 27 '18 at 17:17
$begingroup$
@caffeinemachine How do we know this? M is fixed , A goes to some point other than A'. MA' is symmetric of MA with respect to BC. How is it symmetric with respect to the perpendicular bisector of BC?
$endgroup$
– Lozenges
Dec 27 '18 at 17:25
$begingroup$
Very nice proof.
$endgroup$
– caffeinemachine
Dec 27 '18 at 16:25
$begingroup$
Very nice proof.
$endgroup$
– caffeinemachine
Dec 27 '18 at 16:25
$begingroup$
we are asked to prove angle ABC = angle MIB not angle MIC
$endgroup$
– Lozenges
Dec 27 '18 at 16:33
$begingroup$
we are asked to prove angle ABC = angle MIB not angle MIC
$endgroup$
– Lozenges
Dec 27 '18 at 16:33
$begingroup$
also how do we know MA goes to MA' ?
$endgroup$
– Lozenges
Dec 27 '18 at 16:41
$begingroup$
also how do we know MA goes to MA' ?
$endgroup$
– Lozenges
Dec 27 '18 at 16:41
$begingroup$
@Lozenges What greedoid means is that under the transformation of the plane "reflection about the perpendicular bisector of BC", the line $MA$ transforms to $MA'$.
$endgroup$
– caffeinemachine
Dec 27 '18 at 17:17
$begingroup$
@Lozenges What greedoid means is that under the transformation of the plane "reflection about the perpendicular bisector of BC", the line $MA$ transforms to $MA'$.
$endgroup$
– caffeinemachine
Dec 27 '18 at 17:17
$begingroup$
@caffeinemachine How do we know this? M is fixed , A goes to some point other than A'. MA' is symmetric of MA with respect to BC. How is it symmetric with respect to the perpendicular bisector of BC?
$endgroup$
– Lozenges
Dec 27 '18 at 17:25
$begingroup$
@caffeinemachine How do we know this? M is fixed , A goes to some point other than A'. MA' is symmetric of MA with respect to BC. How is it symmetric with respect to the perpendicular bisector of BC?
$endgroup$
– Lozenges
Dec 27 '18 at 17:25
|
show 2 more comments
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