Mixing
bases of two right triangles inscribed in a semicircle forming the hypotenuse of a larger right triangle?
$begingroup$
I'm unsure why the largest triangle with hypotenuse $r_a + r_b$ must be a right triangle. This is a stepping stone to proving that the area of the 2 smaller circles subtracted from the largest circle is $pi$. The 3 circles are all tangent to one another and the perpendicular line of length 2 is a given.
image
Source: http://datagenetics.com/blog/january62019/index.html
geometry
edited Jan 31 at 20:28
Larry
2,53031131
asked Jan 31 at 19:51
atloo1atloo1
31
$endgroup$
add a comment |
$begingroup$
I'm unsure why the largest triangle with hypotenuse $r_a + r_b$ must be a right triangle. This is a stepping stone to proving that the area of the 2 smaller circles subtracted from the largest circle is $pi$. The 3 circles are all tangent to one another and the perpendicular line of length 2 is a given.
image
Source: http://datagenetics.com/blog/january62019/index.html
geometry
edited Jan 31 at 20:28
Larry
2,53031131
asked Jan 31 at 19:51
atloo1atloo1
31
$endgroup$
1
$begingroup$
I am puzzled by the wording. "Hypotenuse $r_a+ r_b$ must be a right triangle". Only right triangles have hypotenuses!
$endgroup$
– user247327
Jan 31 at 19:58
2
$begingroup$
It's not clear to me exactly what your question is. The large red triangle is a right triangle since it is inscribed in a semicircle. Does that help?
$endgroup$
– rogerl
Jan 31 at 19:59
add a comment |
$begingroup$
I'm unsure why the largest triangle with hypotenuse $r_a + r_b$ must be a right triangle. This is a stepping stone to proving that the area of the 2 smaller circles subtracted from the largest circle is $pi$. The 3 circles are all tangent to one another and the perpendicular line of length 2 is a given.
image
Source: http://datagenetics.com/blog/january62019/index.html
geometry
edited Jan 31 at 20:28
Larry
2,53031131
asked Jan 31 at 19:51
atloo1atloo1
31
$endgroup$
I'm unsure why the largest triangle with hypotenuse $r_a + r_b$ must be a right triangle. This is a stepping stone to proving that the area of the 2 smaller circles subtracted from the largest circle is $pi$. The 3 circles are all tangent to one another and the perpendicular line of length 2 is a given.
image
Source: http://datagenetics.com/blog/january62019/index.html
geometry
geometry
edited Jan 31 at 20:28
Larry
2,53031131
asked Jan 31 at 19:51
atloo1atloo1
31
edited Jan 31 at 20:28
Larry
2,53031131
asked Jan 31 at 19:51
atloo1atloo1
31
edited Jan 31 at 20:28
Larry
2,53031131
edited Jan 31 at 20:28
Larry
2,53031131
edited Jan 31 at 20:28
Larry
2,53031131
2,53031131
asked Jan 31 at 19:51
atloo1atloo1
31
asked Jan 31 at 19:51
atloo1atloo1
31
asked Jan 31 at 19:51
atloo1atloo1
31
31
1
$begingroup$
I am puzzled by the wording. "Hypotenuse $r_a+ r_b$ must be a right triangle". Only right triangles have hypotenuses!
$endgroup$
– user247327
Jan 31 at 19:58
2
$begingroup$
It's not clear to me exactly what your question is. The large red triangle is a right triangle since it is inscribed in a semicircle. Does that help?
$endgroup$
– rogerl
Jan 31 at 19:59
add a comment |
1
$begingroup$
I am puzzled by the wording. "Hypotenuse $r_a+ r_b$ must be a right triangle". Only right triangles have hypotenuses!
$endgroup$
– user247327
Jan 31 at 19:58
2
$begingroup$
It's not clear to me exactly what your question is. The large red triangle is a right triangle since it is inscribed in a semicircle. Does that help?
$endgroup$
– rogerl
Jan 31 at 19:59
1
1
$begingroup$
I am puzzled by the wording. "Hypotenuse $r_a+ r_b$ must be a right triangle". Only right triangles have hypotenuses!
$endgroup$
– user247327
Jan 31 at 19:58
$begingroup$
I am puzzled by the wording. "Hypotenuse $r_a+ r_b$ must be a right triangle". Only right triangles have hypotenuses!
$endgroup$
– user247327
Jan 31 at 19:58
2
2
$begingroup$
It's not clear to me exactly what your question is. The large red triangle is a right triangle since it is inscribed in a semicircle. Does that help?
$endgroup$
– rogerl
Jan 31 at 19:59
$begingroup$
It's not clear to me exactly what your question is. The large red triangle is a right triangle since it is inscribed in a semicircle. Does that help?
$endgroup$
– rogerl
Jan 31 at 19:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is a well-known result that an angle $angle ABC$ will be a right angle if and only if points $A,B,C$ are inscribed in a semicircle with $A$ and $C$ endpoints of its diameter.
Pf: If $A,B,C$ are inscribed in a semicircle with $A$ and $C$ as enpoints of its diameter and $O$ the center of the circle than $AO = BO=CO$ so $triangle AOB$ and $triangle BOC$ are isosceles And $angle OAB cong angle OBA$ and $angle OBC cong angle OCB$. And so as $180 = mangle CAB + mangle ABC + mangle ACB$ and $mangle CAB = mangle OAB =mangle OBA$ and $mangle ABC = mangle OBA + mangle OBC$ and $mangle ACB= mangle OCB=mangle OBC$ we have $180 = 2mangle OBA + 2mangle OCB$. So $mangle ABC = mangle OBA + mangle OCB = 90$.
If $angle ABC$ is a right angle. Than $angle BAC$ and $angle ACB$ are complementary and accute. Construct $M$ on $overline AC$ so that $mangle ABM = mangle MAB$. Then as $angle MBC$ is complimentary to $angle MBA cong angle MAB = angle CAB$ and $angle MCB = angle ACB$ is complimentary to $angle MBC$ have $mangle MCB = mangle MBC$. So $triangle AMB$ and $triangle BMC$ are isosceles and $AM = BM$ and $BM = MC$ so $AM = BM = CM$. So $A,B,C$ are points of a circle centered at $M$. As $A,M, C$ are colinear, $overline{AC}$ is a diameter.
answered Jan 31 at 20:27
fleabloodfleablood
73.9k22891
$endgroup$
add a comment |
$begingroup$
If you have an inscribed angle in a circle, the arc subtended by the angle is twice the angle.
The bottom line is a diameter of the big circle (half of which is shown).
Can you complete the thought from here?
Spoiler:
A diameter cuts a circle into two halves of $pi$ radians. The inscribed angle must be half this, or $pi/2$ radians.
answered Jan 31 at 20:02
JohnJohn
22.8k32550
$endgroup$
add a comment |
$begingroup$
An angle inscribed in a circle will always have a measure of one half of the intercepted arc.
The largest angle in the largest triangle intercepts a $180^circ $ arc (since we know we have three semicircles). Therefore, the inscribed angle opposite this $180^circ $ intercepted arc has half that measure: $90^circ $.
answered Jan 31 at 20:01
Kurt SchwandaKurt Schwanda
40410
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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$begingroup$
It is a well-known result that an angle $angle ABC$ will be a right angle if and only if points $A,B,C$ are inscribed in a semicircle with $A$ and $C$ endpoints of its diameter.
Pf: If $A,B,C$ are inscribed in a semicircle with $A$ and $C$ as enpoints of its diameter and $O$ the center of the circle than $AO = BO=CO$ so $triangle AOB$ and $triangle BOC$ are isosceles And $angle OAB cong angle OBA$ and $angle OBC cong angle OCB$. And so as $180 = mangle CAB + mangle ABC + mangle ACB$ and $mangle CAB = mangle OAB =mangle OBA$ and $mangle ABC = mangle OBA + mangle OBC$ and $mangle ACB= mangle OCB=mangle OBC$ we have $180 = 2mangle OBA + 2mangle OCB$. So $mangle ABC = mangle OBA + mangle OCB = 90$.
If $angle ABC$ is a right angle. Than $angle BAC$ and $angle ACB$ are complementary and accute. Construct $M$ on $overline AC$ so that $mangle ABM = mangle MAB$. Then as $angle MBC$ is complimentary to $angle MBA cong angle MAB = angle CAB$ and $angle MCB = angle ACB$ is complimentary to $angle MBC$ have $mangle MCB = mangle MBC$. So $triangle AMB$ and $triangle BMC$ are isosceles and $AM = BM$ and $BM = MC$ so $AM = BM = CM$. So $A,B,C$ are points of a circle centered at $M$. As $A,M, C$ are colinear, $overline{AC}$ is a diameter.
answered Jan 31 at 20:27
fleabloodfleablood
73.9k22891
$endgroup$
add a comment |
$begingroup$
It is a well-known result that an angle $angle ABC$ will be a right angle if and only if points $A,B,C$ are inscribed in a semicircle with $A$ and $C$ endpoints of its diameter.
Pf: If $A,B,C$ are inscribed in a semicircle with $A$ and $C$ as enpoints of its diameter and $O$ the center of the circle than $AO = BO=CO$ so $triangle AOB$ and $triangle BOC$ are isosceles And $angle OAB cong angle OBA$ and $angle OBC cong angle OCB$. And so as $180 = mangle CAB + mangle ABC + mangle ACB$ and $mangle CAB = mangle OAB =mangle OBA$ and $mangle ABC = mangle OBA + mangle OBC$ and $mangle ACB= mangle OCB=mangle OBC$ we have $180 = 2mangle OBA + 2mangle OCB$. So $mangle ABC = mangle OBA + mangle OCB = 90$.
If $angle ABC$ is a right angle. Than $angle BAC$ and $angle ACB$ are complementary and accute. Construct $M$ on $overline AC$ so that $mangle ABM = mangle MAB$. Then as $angle MBC$ is complimentary to $angle MBA cong angle MAB = angle CAB$ and $angle MCB = angle ACB$ is complimentary to $angle MBC$ have $mangle MCB = mangle MBC$. So $triangle AMB$ and $triangle BMC$ are isosceles and $AM = BM$ and $BM = MC$ so $AM = BM = CM$. So $A,B,C$ are points of a circle centered at $M$. As $A,M, C$ are colinear, $overline{AC}$ is a diameter.
answered Jan 31 at 20:27
fleabloodfleablood
73.9k22891
$endgroup$
add a comment |
$begingroup$
It is a well-known result that an angle $angle ABC$ will be a right angle if and only if points $A,B,C$ are inscribed in a semicircle with $A$ and $C$ endpoints of its diameter.
Pf: If $A,B,C$ are inscribed in a semicircle with $A$ and $C$ as enpoints of its diameter and $O$ the center of the circle than $AO = BO=CO$ so $triangle AOB$ and $triangle BOC$ are isosceles And $angle OAB cong angle OBA$ and $angle OBC cong angle OCB$. And so as $180 = mangle CAB + mangle ABC + mangle ACB$ and $mangle CAB = mangle OAB =mangle OBA$ and $mangle ABC = mangle OBA + mangle OBC$ and $mangle ACB= mangle OCB=mangle OBC$ we have $180 = 2mangle OBA + 2mangle OCB$. So $mangle ABC = mangle OBA + mangle OCB = 90$.
If $angle ABC$ is a right angle. Than $angle BAC$ and $angle ACB$ are complementary and accute. Construct $M$ on $overline AC$ so that $mangle ABM = mangle MAB$. Then as $angle MBC$ is complimentary to $angle MBA cong angle MAB = angle CAB$ and $angle MCB = angle ACB$ is complimentary to $angle MBC$ have $mangle MCB = mangle MBC$. So $triangle AMB$ and $triangle BMC$ are isosceles and $AM = BM$ and $BM = MC$ so $AM = BM = CM$. So $A,B,C$ are points of a circle centered at $M$. As $A,M, C$ are colinear, $overline{AC}$ is a diameter.
answered Jan 31 at 20:27
fleabloodfleablood
73.9k22891
$endgroup$
It is a well-known result that an angle $angle ABC$ will be a right angle if and only if points $A,B,C$ are inscribed in a semicircle with $A$ and $C$ endpoints of its diameter.
Pf: If $A,B,C$ are inscribed in a semicircle with $A$ and $C$ as enpoints of its diameter and $O$ the center of the circle than $AO = BO=CO$ so $triangle AOB$ and $triangle BOC$ are isosceles And $angle OAB cong angle OBA$ and $angle OBC cong angle OCB$. And so as $180 = mangle CAB + mangle ABC + mangle ACB$ and $mangle CAB = mangle OAB =mangle OBA$ and $mangle ABC = mangle OBA + mangle OBC$ and $mangle ACB= mangle OCB=mangle OBC$ we have $180 = 2mangle OBA + 2mangle OCB$. So $mangle ABC = mangle OBA + mangle OCB = 90$.
If $angle ABC$ is a right angle. Than $angle BAC$ and $angle ACB$ are complementary and accute. Construct $M$ on $overline AC$ so that $mangle ABM = mangle MAB$. Then as $angle MBC$ is complimentary to $angle MBA cong angle MAB = angle CAB$ and $angle MCB = angle ACB$ is complimentary to $angle MBC$ have $mangle MCB = mangle MBC$. So $triangle AMB$ and $triangle BMC$ are isosceles and $AM = BM$ and $BM = MC$ so $AM = BM = CM$. So $A,B,C$ are points of a circle centered at $M$. As $A,M, C$ are colinear, $overline{AC}$ is a diameter.
answered Jan 31 at 20:27
fleabloodfleablood
73.9k22891
answered Jan 31 at 20:27
fleabloodfleablood
73.9k22891
answered Jan 31 at 20:27
fleabloodfleablood
73.9k22891
answered Jan 31 at 20:27
fleabloodfleablood
73.9k22891
73.9k22891
add a comment |
add a comment |
$begingroup$
If you have an inscribed angle in a circle, the arc subtended by the angle is twice the angle.
The bottom line is a diameter of the big circle (half of which is shown).
Can you complete the thought from here?
Spoiler:
A diameter cuts a circle into two halves of $pi$ radians. The inscribed angle must be half this, or $pi/2$ radians.
answered Jan 31 at 20:02
JohnJohn
22.8k32550
$endgroup$
add a comment |
$begingroup$
If you have an inscribed angle in a circle, the arc subtended by the angle is twice the angle.
The bottom line is a diameter of the big circle (half of which is shown).
Can you complete the thought from here?
Spoiler:
A diameter cuts a circle into two halves of $pi$ radians. The inscribed angle must be half this, or $pi/2$ radians.
answered Jan 31 at 20:02
JohnJohn
22.8k32550
$endgroup$
add a comment |
$begingroup$
If you have an inscribed angle in a circle, the arc subtended by the angle is twice the angle.
The bottom line is a diameter of the big circle (half of which is shown).
Can you complete the thought from here?
Spoiler:
A diameter cuts a circle into two halves of $pi$ radians. The inscribed angle must be half this, or $pi/2$ radians.
answered Jan 31 at 20:02
JohnJohn
22.8k32550
$endgroup$
If you have an inscribed angle in a circle, the arc subtended by the angle is twice the angle.
The bottom line is a diameter of the big circle (half of which is shown).
Can you complete the thought from here?
Spoiler:
A diameter cuts a circle into two halves of $pi$ radians. The inscribed angle must be half this, or $pi/2$ radians.
answered Jan 31 at 20:02
JohnJohn
22.8k32550
answered Jan 31 at 20:02
JohnJohn
22.8k32550
answered Jan 31 at 20:02
JohnJohn
22.8k32550
answered Jan 31 at 20:02
JohnJohn
22.8k32550
22.8k32550
add a comment |
add a comment |
$begingroup$
An angle inscribed in a circle will always have a measure of one half of the intercepted arc.
The largest angle in the largest triangle intercepts a $180^circ $ arc (since we know we have three semicircles). Therefore, the inscribed angle opposite this $180^circ $ intercepted arc has half that measure: $90^circ $.
answered Jan 31 at 20:01
Kurt SchwandaKurt Schwanda
40410
$endgroup$
add a comment |
$begingroup$
An angle inscribed in a circle will always have a measure of one half of the intercepted arc.
The largest angle in the largest triangle intercepts a $180^circ $ arc (since we know we have three semicircles). Therefore, the inscribed angle opposite this $180^circ $ intercepted arc has half that measure: $90^circ $.
answered Jan 31 at 20:01
Kurt SchwandaKurt Schwanda
40410
$endgroup$
add a comment |
$begingroup$
An angle inscribed in a circle will always have a measure of one half of the intercepted arc.
The largest angle in the largest triangle intercepts a $180^circ $ arc (since we know we have three semicircles). Therefore, the inscribed angle opposite this $180^circ $ intercepted arc has half that measure: $90^circ $.
answered Jan 31 at 20:01
Kurt SchwandaKurt Schwanda
40410
$endgroup$
An angle inscribed in a circle will always have a measure of one half of the intercepted arc.
The largest angle in the largest triangle intercepts a $180^circ $ arc (since we know we have three semicircles). Therefore, the inscribed angle opposite this $180^circ $ intercepted arc has half that measure: $90^circ $.
answered Jan 31 at 20:01
Kurt SchwandaKurt Schwanda
40410
answered Jan 31 at 20:01
Kurt SchwandaKurt Schwanda
40410
answered Jan 31 at 20:01
Kurt SchwandaKurt Schwanda
40410
answered Jan 31 at 20:01
Kurt SchwandaKurt Schwanda
40410
40410
add a comment |
add a comment |
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$begingroup$
I am puzzled by the wording. "Hypotenuse $r_a+ r_b$ must be a right triangle". Only right triangles have hypotenuses!
$endgroup$
– user247327
Jan 31 at 19:58
2
$begingroup$
It's not clear to me exactly what your question is. The large red triangle is a right triangle since it is inscribed in a semicircle. Does that help?
$endgroup$
– rogerl
Jan 31 at 19:59