High school geometry problem: Reflect a vertex about opposite side.












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$begingroup$


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: $angle ABC=angle MIB$.




enter image description here



I am not able to make any progress.










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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: $angle ABC=angle MIB$.




    enter image description here



    I am not able to make any progress.










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      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: $angle ABC=angle MIB$.




      enter image description here



      I am not able to make any progress.










      share|cite|improve this question











      $endgroup$




      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: $angle ABC=angle MIB$.




      enter image description here



      I am not able to make any progress.







      geometry contest-math euclidean-geometry reflection geometric-transformation






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 18:41









      greedoid

      38.7k114797




      38.7k114797










      asked Dec 27 '18 at 14:49









      caffeinemachinecaffeinemachine

      6,50821350




      6,50821350






















          1 Answer
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          active

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          4












          $begingroup$

          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 $$angle ABC = angle AJC = angle MJC = angle MIB$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes









          4












          $begingroup$

          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 $$angle ABC = angle AJC = angle MJC = angle MIB$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          4












          $begingroup$

          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 $$angle ABC = angle AJC = angle MJC = angle MIB$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          4












          4








          4





          $begingroup$

          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 $$angle ABC = angle AJC = angle MJC = angle MIB$$






          share|cite|improve this answer











          $endgroup$



          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 $$angle ABC = angle AJC = angle MJC = angle MIB$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 27 '18 at 16:42

























          answered Dec 27 '18 at 15:24









          greedoidgreedoid

          38.7k114797




          38.7k114797












          • $begingroup$
            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










          • $begingroup$
            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


















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