Show that $CDperp AB $.












2














Let $triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.



If $angle BAA_1=30°$ and $Din [AB] $ s.t. $CD=AB $ show that $CDperp AB $.



My idea : I draw $A_1Tperp AB $, $Tin [AB]$.



I have to show that $A_1T $ is the middle line in $triangle CBD $ $Leftrightarrow$ $A_1T=frac {CD}{2}=frac {AB}{2} $.



But $A_1T=frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.






geometry euclidean-geometry circle







share|cite|improve this question
























  • Is this supposed to be true for, for example, an equilateral triangle?
    – random
    14 hours ago










  • @random In an equilateral triangle there is no point $D $ with that property
    – rafa
    13 hours ago
















2














Let $triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.



If $angle BAA_1=30°$ and $Din [AB] $ s.t. $CD=AB $ show that $CDperp AB $.



My idea : I draw $A_1Tperp AB $, $Tin [AB]$.



I have to show that $A_1T $ is the middle line in $triangle CBD $ $Leftrightarrow$ $A_1T=frac {CD}{2}=frac {AB}{2} $.



But $A_1T=frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.






geometry euclidean-geometry circle







share|cite|improve this question
























  • Is this supposed to be true for, for example, an equilateral triangle?
    – random
    14 hours ago










  • @random In an equilateral triangle there is no point $D $ with that property
    – rafa
    13 hours ago














2












2








2


2





Let $triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.



If $angle BAA_1=30°$ and $Din [AB] $ s.t. $CD=AB $ show that $CDperp AB $.



My idea : I draw $A_1Tperp AB $, $Tin [AB]$.



I have to show that $A_1T $ is the middle line in $triangle CBD $ $Leftrightarrow$ $A_1T=frac {CD}{2}=frac {AB}{2} $.



But $A_1T=frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.






geometry euclidean-geometry circle







share|cite|improve this question















Let $triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.



If $angle BAA_1=30°$ and $Din [AB] $ s.t. $CD=AB $ show that $CDperp AB $.



My idea : I draw $A_1Tperp AB $, $Tin [AB]$.



I have to show that $A_1T $ is the middle line in $triangle CBD $ $Leftrightarrow$ $A_1T=frac {CD}{2}=frac {AB}{2} $.



But $A_1T=frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.






geometry euclidean-geometry circle




geometry euclidean-geometry circle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 14 hours ago

























asked 15 hours ago









rafa

583212




583212












  • Is this supposed to be true for, for example, an equilateral triangle?
    – random
    14 hours ago










  • @random In an equilateral triangle there is no point $D $ with that property
    – rafa
    13 hours ago


















  • Is this supposed to be true for, for example, an equilateral triangle?
    – random
    14 hours ago










  • @random In an equilateral triangle there is no point $D $ with that property
    – rafa
    13 hours ago
















Is this supposed to be true for, for example, an equilateral triangle?
– random
14 hours ago




Is this supposed to be true for, for example, an equilateral triangle?
– random
14 hours ago












@random In an equilateral triangle there is no point $D $ with that property
– rafa
13 hours ago




@random In an equilateral triangle there is no point $D $ with that property
– rafa
13 hours ago










1 Answer
1






active

oldest

votes


















1














Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.



Then $A_1 = left(1 - frac1{sqrt3},frac13right).$



Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$



The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$



Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$



We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$



Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$



In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$



It seems some necessary condition is missing from your problem statement.






share|cite|improve this answer























  • I came to the same conclusion by drawing a counterexample.
    – Oldboy
    12 hours ago











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1 Answer
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1 Answer
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active

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1














Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.



Then $A_1 = left(1 - frac1{sqrt3},frac13right).$



Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$



The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$



Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$



We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$



Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$



In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$



It seems some necessary condition is missing from your problem statement.






share|cite|improve this answer























  • I came to the same conclusion by drawing a counterexample.
    – Oldboy
    12 hours ago
















1














Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.



Then $A_1 = left(1 - frac1{sqrt3},frac13right).$



Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$



The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$



Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$



We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$



Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$



In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$



It seems some necessary condition is missing from your problem statement.






share|cite|improve this answer























  • I came to the same conclusion by drawing a counterexample.
    – Oldboy
    12 hours ago














1












1








1






Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.



Then $A_1 = left(1 - frac1{sqrt3},frac13right).$



Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$



The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$



Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$



We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$



Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$



In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$



It seems some necessary condition is missing from your problem statement.






share|cite|improve this answer














Let $A = (1,0),$ $B=(0,0),$ and $C = left(2 - frac2{sqrt3},frac23right).$
Since $2 - frac2{sqrt3} approx 0.845299,$
angles $angle CAB$ and $angle CBA$ are acute,
and since the distance from $C$ to $left(frac12,0right)$ is greater than
$frac12,$ the angle $angle ABC$ also is acute.



Then $A_1 = left(1 - frac1{sqrt3},frac13right).$



Line $AA_1$ has the equation $y = frac1{sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-frac1{sqrt3} = tan(-30^circ),$
so $angle BAA_1 = 30^circ.$



The circle of radius $AB = 1$ around $C$ has equation
$$left(x + frac2{sqrt3} - 2right)^2 + left(y - frac23right)^2 = 1.$$



Put $y = 0$ in this equation and we find that
$$left(x + frac2{sqrt3} - 2right)^2 + frac49 = 1,$$
$$x + frac2{sqrt3} - 2 = pm sqrt{frac59},$$
$$x = 2 - frac2{sqrt3} pm frac{sqrt5}3.$$



We have $frac{sqrt5}3 approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - frac2{sqrt3} - frac{sqrt5}3 approx 0.099943.$



Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$



In fact, if $C$ is any point $(x,y)$ that satisfies $y = frac1{sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$angle BAA_1 = 30^circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD perp AB$
is if $C = left(1 - frac{sqrt3}2,1right).$



It seems some necessary condition is missing from your problem statement.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 11 hours ago

























answered 12 hours ago









David K

52.7k340115




52.7k340115












  • I came to the same conclusion by drawing a counterexample.
    – Oldboy
    12 hours ago


















  • I came to the same conclusion by drawing a counterexample.
    – Oldboy
    12 hours ago
















I came to the same conclusion by drawing a counterexample.
– Oldboy
12 hours ago




I came to the same conclusion by drawing a counterexample.
– Oldboy
12 hours ago


















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