Proof regarding altitudes of a triangle and a midpoint of one of its sides












0












$begingroup$


Let $triangle ABC$ be a triangle with altitudes $overline{AH}$ and $overline{BK}$. Consider the axis of the segment $ overline{HK}$. Let $M$ be the point of intersection between the axis and the side $overline{AB}$.



I have to prove that $overline{AM} cong overline{MB}$.



If the triangle $triangle ABC$ is isosceles the the proof is easy; but in the general case I don't know how to proceed. I have noticed that $overline{MH} cong overline{MK}$ but I can't conclude.



Thank you for your help!






euclidean-geometry triangle plane-geometry







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$endgroup$












  • $begingroup$
    What do you mean by the axis of segment $overline {HK}$?
    $endgroup$
    – Dr. Mathva
    Dec 16 '18 at 11:12










  • $begingroup$
    Is it the perpendicular bisector of $overline {HK}$?
    $endgroup$
    – Dr. Mathva
    Dec 16 '18 at 11:13










  • $begingroup$
    Yes, it is. Take the midpoint of $overline{HK}$ and consider the straight line passing through this point and perpendicul to $overline{HK}$.Thank you for your time and for your clarification
    $endgroup$
    – Hermione
    Dec 16 '18 at 11:20


















0












$begingroup$


Let $triangle ABC$ be a triangle with altitudes $overline{AH}$ and $overline{BK}$. Consider the axis of the segment $ overline{HK}$. Let $M$ be the point of intersection between the axis and the side $overline{AB}$.



I have to prove that $overline{AM} cong overline{MB}$.



If the triangle $triangle ABC$ is isosceles the the proof is easy; but in the general case I don't know how to proceed. I have noticed that $overline{MH} cong overline{MK}$ but I can't conclude.



Thank you for your help!






euclidean-geometry triangle plane-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by the axis of segment $overline {HK}$?
    $endgroup$
    – Dr. Mathva
    Dec 16 '18 at 11:12










  • $begingroup$
    Is it the perpendicular bisector of $overline {HK}$?
    $endgroup$
    – Dr. Mathva
    Dec 16 '18 at 11:13










  • $begingroup$
    Yes, it is. Take the midpoint of $overline{HK}$ and consider the straight line passing through this point and perpendicul to $overline{HK}$.Thank you for your time and for your clarification
    $endgroup$
    – Hermione
    Dec 16 '18 at 11:20
















0












0








0





$begingroup$


Let $triangle ABC$ be a triangle with altitudes $overline{AH}$ and $overline{BK}$. Consider the axis of the segment $ overline{HK}$. Let $M$ be the point of intersection between the axis and the side $overline{AB}$.



I have to prove that $overline{AM} cong overline{MB}$.



If the triangle $triangle ABC$ is isosceles the the proof is easy; but in the general case I don't know how to proceed. I have noticed that $overline{MH} cong overline{MK}$ but I can't conclude.



Thank you for your help!






euclidean-geometry triangle plane-geometry







share|cite|improve this question











$endgroup$




Let $triangle ABC$ be a triangle with altitudes $overline{AH}$ and $overline{BK}$. Consider the axis of the segment $ overline{HK}$. Let $M$ be the point of intersection between the axis and the side $overline{AB}$.



I have to prove that $overline{AM} cong overline{MB}$.



If the triangle $triangle ABC$ is isosceles the the proof is easy; but in the general case I don't know how to proceed. I have noticed that $overline{MH} cong overline{MK}$ but I can't conclude.



Thank you for your help!






euclidean-geometry triangle plane-geometry




euclidean-geometry triangle plane-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 11:20









Bernard

121k740116




121k740116










asked Dec 16 '18 at 10:26









HermioneHermione

19619




19619












  • $begingroup$
    What do you mean by the axis of segment $overline {HK}$?
    $endgroup$
    – Dr. Mathva
    Dec 16 '18 at 11:12










  • $begingroup$
    Is it the perpendicular bisector of $overline {HK}$?
    $endgroup$
    – Dr. Mathva
    Dec 16 '18 at 11:13










  • $begingroup$
    Yes, it is. Take the midpoint of $overline{HK}$ and consider the straight line passing through this point and perpendicul to $overline{HK}$.Thank you for your time and for your clarification
    $endgroup$
    – Hermione
    Dec 16 '18 at 11:20




















  • $begingroup$
    What do you mean by the axis of segment $overline {HK}$?
    $endgroup$
    – Dr. Mathva
    Dec 16 '18 at 11:12










  • $begingroup$
    Is it the perpendicular bisector of $overline {HK}$?
    $endgroup$
    – Dr. Mathva
    Dec 16 '18 at 11:13










  • $begingroup$
    Yes, it is. Take the midpoint of $overline{HK}$ and consider the straight line passing through this point and perpendicul to $overline{HK}$.Thank you for your time and for your clarification
    $endgroup$
    – Hermione
    Dec 16 '18 at 11:20


















$begingroup$
What do you mean by the axis of segment $overline {HK}$?
$endgroup$
– Dr. Mathva
Dec 16 '18 at 11:12




$begingroup$
What do you mean by the axis of segment $overline {HK}$?
$endgroup$
– Dr. Mathva
Dec 16 '18 at 11:12












$begingroup$
Is it the perpendicular bisector of $overline {HK}$?
$endgroup$
– Dr. Mathva
Dec 16 '18 at 11:13




$begingroup$
Is it the perpendicular bisector of $overline {HK}$?
$endgroup$
– Dr. Mathva
Dec 16 '18 at 11:13












$begingroup$
Yes, it is. Take the midpoint of $overline{HK}$ and consider the straight line passing through this point and perpendicul to $overline{HK}$.Thank you for your time and for your clarification
$endgroup$
– Hermione
Dec 16 '18 at 11:20






$begingroup$
Yes, it is. Take the midpoint of $overline{HK}$ and consider the straight line passing through this point and perpendicul to $overline{HK}$.Thank you for your time and for your clarification
$endgroup$
– Hermione
Dec 16 '18 at 11:20












2 Answers
2






active

oldest

votes


















1












$begingroup$

Notice that both triangles $Delta ABH$ and $Delta BKA$ have a right angle $(=90°)$ and share the basis $overline {AB}$.



By the inversion of Thales's Theorem, they are inscribed in the same circle $omega$ with diameter $overline {AB}$.



$overline {KH}$ is, therefore, one chord of $omega$, implying that the perpendicular bisector of $overline {KH}$ passes through the circle's center, i.e. $overline {AB}$'s midpoint $M$, see why?




Since $overline {AB}$ is the diameter, the midpoint $M_{AB}$ of $overline {AB}$ is $omega$'s centre. Notice now that points $H$ and $K$ both lie on the circumference. Since $$overline {M_{AB}K}=overline {M_{AB}H}=omega text{'s radius}$$ which implies that $M_{AB}$ lies on the perpendicular bisector of $overline {HK}$




It follows that




$$overline {AM}cong overline {MB}$$







share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Since $$angle AKB = angle AHB = 90$$
    we see that $A,B,H,K$ are concyclic. The circle around $A,B,H,K$ has diameter $AB$ (by Thales theorem) and thus a conclusion.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Notice that both triangles $Delta ABH$ and $Delta BKA$ have a right angle $(=90°)$ and share the basis $overline {AB}$.



      By the inversion of Thales's Theorem, they are inscribed in the same circle $omega$ with diameter $overline {AB}$.



      $overline {KH}$ is, therefore, one chord of $omega$, implying that the perpendicular bisector of $overline {KH}$ passes through the circle's center, i.e. $overline {AB}$'s midpoint $M$, see why?




      Since $overline {AB}$ is the diameter, the midpoint $M_{AB}$ of $overline {AB}$ is $omega$'s centre. Notice now that points $H$ and $K$ both lie on the circumference. Since $$overline {M_{AB}K}=overline {M_{AB}H}=omega text{'s radius}$$ which implies that $M_{AB}$ lies on the perpendicular bisector of $overline {HK}$




      It follows that




      $$overline {AM}cong overline {MB}$$







      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Notice that both triangles $Delta ABH$ and $Delta BKA$ have a right angle $(=90°)$ and share the basis $overline {AB}$.



        By the inversion of Thales's Theorem, they are inscribed in the same circle $omega$ with diameter $overline {AB}$.



        $overline {KH}$ is, therefore, one chord of $omega$, implying that the perpendicular bisector of $overline {KH}$ passes through the circle's center, i.e. $overline {AB}$'s midpoint $M$, see why?




        Since $overline {AB}$ is the diameter, the midpoint $M_{AB}$ of $overline {AB}$ is $omega$'s centre. Notice now that points $H$ and $K$ both lie on the circumference. Since $$overline {M_{AB}K}=overline {M_{AB}H}=omega text{'s radius}$$ which implies that $M_{AB}$ lies on the perpendicular bisector of $overline {HK}$




        It follows that




        $$overline {AM}cong overline {MB}$$







        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Notice that both triangles $Delta ABH$ and $Delta BKA$ have a right angle $(=90°)$ and share the basis $overline {AB}$.



          By the inversion of Thales's Theorem, they are inscribed in the same circle $omega$ with diameter $overline {AB}$.



          $overline {KH}$ is, therefore, one chord of $omega$, implying that the perpendicular bisector of $overline {KH}$ passes through the circle's center, i.e. $overline {AB}$'s midpoint $M$, see why?




          Since $overline {AB}$ is the diameter, the midpoint $M_{AB}$ of $overline {AB}$ is $omega$'s centre. Notice now that points $H$ and $K$ both lie on the circumference. Since $$overline {M_{AB}K}=overline {M_{AB}H}=omega text{'s radius}$$ which implies that $M_{AB}$ lies on the perpendicular bisector of $overline {HK}$




          It follows that




          $$overline {AM}cong overline {MB}$$







          share|cite|improve this answer











          $endgroup$



          Notice that both triangles $Delta ABH$ and $Delta BKA$ have a right angle $(=90°)$ and share the basis $overline {AB}$.



          By the inversion of Thales's Theorem, they are inscribed in the same circle $omega$ with diameter $overline {AB}$.



          $overline {KH}$ is, therefore, one chord of $omega$, implying that the perpendicular bisector of $overline {KH}$ passes through the circle's center, i.e. $overline {AB}$'s midpoint $M$, see why?




          Since $overline {AB}$ is the diameter, the midpoint $M_{AB}$ of $overline {AB}$ is $omega$'s centre. Notice now that points $H$ and $K$ both lie on the circumference. Since $$overline {M_{AB}K}=overline {M_{AB}H}=omega text{'s radius}$$ which implies that $M_{AB}$ lies on the perpendicular bisector of $overline {HK}$




          It follows that




          $$overline {AM}cong overline {MB}$$








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 11:36

























          answered Dec 16 '18 at 11:28









          Dr. MathvaDr. Mathva

          1,255318




          1,255318























              1












              $begingroup$

              Since $$angle AKB = angle AHB = 90$$
              we see that $A,B,H,K$ are concyclic. The circle around $A,B,H,K$ has diameter $AB$ (by Thales theorem) and thus a conclusion.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Since $$angle AKB = angle AHB = 90$$
                we see that $A,B,H,K$ are concyclic. The circle around $A,B,H,K$ has diameter $AB$ (by Thales theorem) and thus a conclusion.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Since $$angle AKB = angle AHB = 90$$
                  we see that $A,B,H,K$ are concyclic. The circle around $A,B,H,K$ has diameter $AB$ (by Thales theorem) and thus a conclusion.






                  share|cite|improve this answer









                  $endgroup$



                  Since $$angle AKB = angle AHB = 90$$
                  we see that $A,B,H,K$ are concyclic. The circle around $A,B,H,K$ has diameter $AB$ (by Thales theorem) and thus a conclusion.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 11:48









                  greedoidgreedoid

                  41.9k1152105




                  41.9k1152105






























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