Congruent triangles in 3 tangent circle configurationThree points on a lineKorean Math Olympiad 2005 (trapezoid & tangent circles)Problem about circle tangentsThree mutually-tangent circles have centers at given distances from each other; find each radius, and find the area between the circlesCenter of Soddy CircleProve parallel line is tangent to second circleFind the length of a tangent line of a circletangent circles radical axis proofAre these triangles congruent?Ceva's Theorem: Proving lines in a specifically constructed triangle intersect
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Congruent triangles in 3 tangent circle configuration
Three points on a lineKorean Math Olympiad 2005 (trapezoid & tangent circles)Problem about circle tangentsThree mutually-tangent circles have centers at given distances from each other; find each radius, and find the area between the circlesCenter of Soddy CircleProve parallel line is tangent to second circleFind the length of a tangent line of a circletangent circles radical axis proofAre these triangles congruent?Ceva's Theorem: Proving lines in a specifically constructed triangle intersect
$begingroup$
Two circles $mathcalC_1$ and $mathcalC_2$ of centers $O_1$ and $O_2$ are externally tangent at $I$ and internally tangent to a third circle $mathcalC$ of center $O$ that is colinear with $O_1$ and $O_2$ as depicted below.
A line going through $I$ intersects the three circles at the points $A, B, C, D$ (see figure below).
How to prove that $AB=CD$ without using trigonometry?
I tried to show that the triangles $Delta ABO$ and $Delta DCO$ are congruent, but I was unable to get the needed angle equalities - $OA=OD$ and $angle A=angle D$ are obvious.
geometry euclidean-geometry
asked Mar 21 at 19:43
user2471user2471
604
$endgroup$
|
show 2 more comments
$begingroup$
Two circles $mathcalC_1$ and $mathcalC_2$ of centers $O_1$ and $O_2$ are externally tangent at $I$ and internally tangent to a third circle $mathcalC$ of center $O$ that is colinear with $O_1$ and $O_2$ as depicted below.
A line going through $I$ intersects the three circles at the points $A, B, C, D$ (see figure below).
How to prove that $AB=CD$ without using trigonometry?
I tried to show that the triangles $Delta ABO$ and $Delta DCO$ are congruent, but I was unable to get the needed angle equalities - $OA=OD$ and $angle A=angle D$ are obvious.
geometry euclidean-geometry
asked Mar 21 at 19:43
user2471user2471
604
$endgroup$
$begingroup$
In your figure $O$ lies on line $O_1O_2$. Is this an additional hypothesis?
$endgroup$
– Matteo
Mar 21 at 22:07
1
$begingroup$
Yes! I forgot that
$endgroup$
– user2471
Mar 21 at 22:21
$begingroup$
that explains the counter example below...
$endgroup$
– Matteo
Mar 21 at 22:22
$begingroup$
@Matteo my Apologies!
$endgroup$
– user2471
Mar 21 at 22:26
$begingroup$
If you name the endpoints of the green diameter say M and N, then $angle MBI$ and $angle NCI$ are both right, hence right triangles $triangle MBI$ and $triangle NCI$ are similar (with the scale equal to the ratio of radii of small circles); then $I$ divides $BC$ in the same proportion as it divides the green diameter $MN$. Alas, I can't see yet how it can help...
$endgroup$
– CiaPan
Mar 21 at 22:52
|
show 2 more comments
$begingroup$
Two circles $mathcalC_1$ and $mathcalC_2$ of centers $O_1$ and $O_2$ are externally tangent at $I$ and internally tangent to a third circle $mathcalC$ of center $O$ that is colinear with $O_1$ and $O_2$ as depicted below.
A line going through $I$ intersects the three circles at the points $A, B, C, D$ (see figure below).
How to prove that $AB=CD$ without using trigonometry?
I tried to show that the triangles $Delta ABO$ and $Delta DCO$ are congruent, but I was unable to get the needed angle equalities - $OA=OD$ and $angle A=angle D$ are obvious.
geometry euclidean-geometry
asked Mar 21 at 19:43
user2471user2471
604
$endgroup$
Two circles $mathcalC_1$ and $mathcalC_2$ of centers $O_1$ and $O_2$ are externally tangent at $I$ and internally tangent to a third circle $mathcalC$ of center $O$ that is colinear with $O_1$ and $O_2$ as depicted below.
A line going through $I$ intersects the three circles at the points $A, B, C, D$ (see figure below).
How to prove that $AB=CD$ without using trigonometry?
I tried to show that the triangles $Delta ABO$ and $Delta DCO$ are congruent, but I was unable to get the needed angle equalities - $OA=OD$ and $angle A=angle D$ are obvious.
geometry euclidean-geometry
geometry euclidean-geometry
asked Mar 21 at 19:43
user2471user2471
604
asked Mar 21 at 19:43
user2471user2471
604
edited Mar 21 at 22:23
user2471
asked Mar 21 at 19:43
user2471user2471
604
asked Mar 21 at 19:43
user2471user2471
604
asked Mar 21 at 19:43
user2471user2471
604
604
$begingroup$
In your figure $O$ lies on line $O_1O_2$. Is this an additional hypothesis?
$endgroup$
– Matteo
Mar 21 at 22:07
1
$begingroup$
Yes! I forgot that
$endgroup$
– user2471
Mar 21 at 22:21
$begingroup$
that explains the counter example below...
$endgroup$
– Matteo
Mar 21 at 22:22
$begingroup$
@Matteo my Apologies!
$endgroup$
– user2471
Mar 21 at 22:26
$begingroup$
If you name the endpoints of the green diameter say M and N, then $angle MBI$ and $angle NCI$ are both right, hence right triangles $triangle MBI$ and $triangle NCI$ are similar (with the scale equal to the ratio of radii of small circles); then $I$ divides $BC$ in the same proportion as it divides the green diameter $MN$. Alas, I can't see yet how it can help...
$endgroup$
– CiaPan
Mar 21 at 22:52
|
show 2 more comments
$begingroup$
In your figure $O$ lies on line $O_1O_2$. Is this an additional hypothesis?
$endgroup$
– Matteo
Mar 21 at 22:07
1
$begingroup$
Yes! I forgot that
$endgroup$
– user2471
Mar 21 at 22:21
$begingroup$
that explains the counter example below...
$endgroup$
– Matteo
Mar 21 at 22:22
$begingroup$
@Matteo my Apologies!
$endgroup$
– user2471
Mar 21 at 22:26
$begingroup$
If you name the endpoints of the green diameter say M and N, then $angle MBI$ and $angle NCI$ are both right, hence right triangles $triangle MBI$ and $triangle NCI$ are similar (with the scale equal to the ratio of radii of small circles); then $I$ divides $BC$ in the same proportion as it divides the green diameter $MN$. Alas, I can't see yet how it can help...
$endgroup$
– CiaPan
Mar 21 at 22:52
$begingroup$
In your figure $O$ lies on line $O_1O_2$. Is this an additional hypothesis?
$endgroup$
– Matteo
Mar 21 at 22:07
$begingroup$
In your figure $O$ lies on line $O_1O_2$. Is this an additional hypothesis?
$endgroup$
– Matteo
Mar 21 at 22:07
1
1
$begingroup$
Yes! I forgot that
$endgroup$
– user2471
Mar 21 at 22:21
$begingroup$
Yes! I forgot that
$endgroup$
– user2471
Mar 21 at 22:21
$begingroup$
that explains the counter example below...
$endgroup$
– Matteo
Mar 21 at 22:22
$begingroup$
that explains the counter example below...
$endgroup$
– Matteo
Mar 21 at 22:22
$begingroup$
@Matteo my Apologies!
$endgroup$
– user2471
Mar 21 at 22:26
$begingroup$
@Matteo my Apologies!
$endgroup$
– user2471
Mar 21 at 22:26
$begingroup$
If you name the endpoints of the green diameter say M and N, then $angle MBI$ and $angle NCI$ are both right, hence right triangles $triangle MBI$ and $triangle NCI$ are similar (with the scale equal to the ratio of radii of small circles); then $I$ divides $BC$ in the same proportion as it divides the green diameter $MN$. Alas, I can't see yet how it can help...
$endgroup$
– CiaPan
Mar 21 at 22:52
$begingroup$
If you name the endpoints of the green diameter say M and N, then $angle MBI$ and $angle NCI$ are both right, hence right triangles $triangle MBI$ and $triangle NCI$ are similar (with the scale equal to the ratio of radii of small circles); then $I$ divides $BC$ in the same proportion as it divides the green diameter $MN$. Alas, I can't see yet how it can help...
$endgroup$
– CiaPan
Mar 21 at 22:52
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
After OP added the assumption of collinearity of all three circle centers here is the answer.
Let's draw lines from points M, N, O perpendicular to the chord AD.
Triangles MBI, OSI and NCI are all right and similar. Then S is the midpoint of AD and the lengths between B, I, S and C keep the same proportion as the lengths between M, I, O and N, respectively (they are actually a parallel projection between the lines MN and BC). Hence S is not only a midpoint of AD, but also a midpoint of BC. As a result
$$AB = AS - BS = frac 12 AD - frac 12 BC = DS - CS = DC$$
Q.E.D.
answered Mar 21 at 23:21
CiaPanCiaPan
10.3k11248
$endgroup$
add a comment |
$begingroup$
Something is wrong. I'm afraid that without some additional assumptions regarding the 'line through I' (or maybe the circles' configuration) your claim is unprovable...
answered Mar 21 at 21:55
CiaPanCiaPan
10.3k11248
$endgroup$
$begingroup$
My apologies! I forgot one crucial assumption: colinearity of the circle centers!
$endgroup$
– user2471
Mar 21 at 22:28
add a comment |
$begingroup$
I will assume that points $O,O_1,O_2$ are collinear in accordance with your drawing. Otherwise the claim is in general not valid.
Let the points of intersecttion of circle $mathcalC$ with circles $mathcalC_1$ and $mathcalC_2$ be $I_1$ and $I_2$, respectively. Let $angle O_1IA=angle O_2IB$ be $alpha$. Let $R_1$ and $R_2$ be the radii of $mathcalC_1$ and $mathcalC_2$, respectively.
We have
$$
IO=R_2-R_1,quad IB=2R_1cosalpha,quad IC=2R_2cosalpha.
$$
The last two equalities follow from considering the right triangles $IBI_1$ and $ICI_2$.
By the law of cosines one then obtains:
$$beginalign
OB^2&=(R_2-R_1)^2+(2R_1cosalpha)^2+2(R_2-R_1)(2R_1cosalpha)cosalpha
=IO^2+IBcdot IC,\
OC^2&=(R_2-R_1)^2+(2R_2cosalpha)^2-2(R_2-R_1)(2R_2cosalpha)cosalpha
=IO^2+IBcdot IC.
endalign
$$
Thus, $OB=OC$. The rest is simple.
answered Mar 21 at 21:29
useruser
6,21911031
$endgroup$
$begingroup$
OP requires not to use trigonometry.
$endgroup$
– Matteo
Mar 21 at 22:25
$begingroup$
@Matteo I have noticed it. I however do not know if it refers to the usage of cosine law (or Pythagorean theorem). No other "trigonometric" properties are used. In fact $cosalpha$ can be considered here as an abbreviation for $fracIB2R_1=fracIC2R_2$.
$endgroup$
– user
Mar 21 at 22:47
add a comment |
$begingroup$
Here is a possibile path.
$angle O_1BI congangle O_2CI$ (can you tell why?). Therefore $O_1Bparallel O_2C$.- Consequently $angle BO_1O cong angle CO_2O$.
$triangle O_1BO cong triangle O_2CO$ (can you tell why?).- In particular, $OBcong OC$. Thus $triangle OBC$ is isosceles.
And the thesis follows from what you already noticed, since now you can demonstrate that $triangle ABO cong triangle CDO$, as you correctly wished to show.
answered Mar 21 at 23:01
MatteoMatteo
1,302313
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
After OP added the assumption of collinearity of all three circle centers here is the answer.
Let's draw lines from points M, N, O perpendicular to the chord AD.
Triangles MBI, OSI and NCI are all right and similar. Then S is the midpoint of AD and the lengths between B, I, S and C keep the same proportion as the lengths between M, I, O and N, respectively (they are actually a parallel projection between the lines MN and BC). Hence S is not only a midpoint of AD, but also a midpoint of BC. As a result
$$AB = AS - BS = frac 12 AD - frac 12 BC = DS - CS = DC$$
Q.E.D.
answered Mar 21 at 23:21
CiaPanCiaPan
10.3k11248
$endgroup$
add a comment |
$begingroup$
After OP added the assumption of collinearity of all three circle centers here is the answer.
Let's draw lines from points M, N, O perpendicular to the chord AD.
Triangles MBI, OSI and NCI are all right and similar. Then S is the midpoint of AD and the lengths between B, I, S and C keep the same proportion as the lengths between M, I, O and N, respectively (they are actually a parallel projection between the lines MN and BC). Hence S is not only a midpoint of AD, but also a midpoint of BC. As a result
$$AB = AS - BS = frac 12 AD - frac 12 BC = DS - CS = DC$$
Q.E.D.
answered Mar 21 at 23:21
CiaPanCiaPan
10.3k11248
$endgroup$
add a comment |
$begingroup$
After OP added the assumption of collinearity of all three circle centers here is the answer.
Let's draw lines from points M, N, O perpendicular to the chord AD.
Triangles MBI, OSI and NCI are all right and similar. Then S is the midpoint of AD and the lengths between B, I, S and C keep the same proportion as the lengths between M, I, O and N, respectively (they are actually a parallel projection between the lines MN and BC). Hence S is not only a midpoint of AD, but also a midpoint of BC. As a result
$$AB = AS - BS = frac 12 AD - frac 12 BC = DS - CS = DC$$
Q.E.D.
answered Mar 21 at 23:21
CiaPanCiaPan
10.3k11248
$endgroup$
After OP added the assumption of collinearity of all three circle centers here is the answer.
Let's draw lines from points M, N, O perpendicular to the chord AD.
Triangles MBI, OSI and NCI are all right and similar. Then S is the midpoint of AD and the lengths between B, I, S and C keep the same proportion as the lengths between M, I, O and N, respectively (they are actually a parallel projection between the lines MN and BC). Hence S is not only a midpoint of AD, but also a midpoint of BC. As a result
$$AB = AS - BS = frac 12 AD - frac 12 BC = DS - CS = DC$$
Q.E.D.
answered Mar 21 at 23:21
CiaPanCiaPan
10.3k11248
edited Mar 24 at 18:20
answered Mar 21 at 23:21
CiaPanCiaPan
10.3k11248
answered Mar 21 at 23:21
CiaPanCiaPan
10.3k11248
answered Mar 21 at 23:21
CiaPanCiaPan
10.3k11248
10.3k11248
add a comment |
add a comment |
$begingroup$
Something is wrong. I'm afraid that without some additional assumptions regarding the 'line through I' (or maybe the circles' configuration) your claim is unprovable...
answered Mar 21 at 21:55
CiaPanCiaPan
10.3k11248
$endgroup$
$begingroup$
My apologies! I forgot one crucial assumption: colinearity of the circle centers!
$endgroup$
– user2471
Mar 21 at 22:28
add a comment |
$begingroup$
Something is wrong. I'm afraid that without some additional assumptions regarding the 'line through I' (or maybe the circles' configuration) your claim is unprovable...
answered Mar 21 at 21:55
CiaPanCiaPan
10.3k11248
$endgroup$
$begingroup$
My apologies! I forgot one crucial assumption: colinearity of the circle centers!
$endgroup$
– user2471
Mar 21 at 22:28
add a comment |
$begingroup$
Something is wrong. I'm afraid that without some additional assumptions regarding the 'line through I' (or maybe the circles' configuration) your claim is unprovable...
answered Mar 21 at 21:55
CiaPanCiaPan
10.3k11248
$endgroup$
Something is wrong. I'm afraid that without some additional assumptions regarding the 'line through I' (or maybe the circles' configuration) your claim is unprovable...
answered Mar 21 at 21:55
CiaPanCiaPan
10.3k11248
answered Mar 21 at 21:55
CiaPanCiaPan
10.3k11248
answered Mar 21 at 21:55
CiaPanCiaPan
10.3k11248
answered Mar 21 at 21:55
CiaPanCiaPan
10.3k11248
10.3k11248
$begingroup$
My apologies! I forgot one crucial assumption: colinearity of the circle centers!
$endgroup$
– user2471
Mar 21 at 22:28
add a comment |
$begingroup$
My apologies! I forgot one crucial assumption: colinearity of the circle centers!
$endgroup$
– user2471
Mar 21 at 22:28
$begingroup$
My apologies! I forgot one crucial assumption: colinearity of the circle centers!
$endgroup$
– user2471
Mar 21 at 22:28
$begingroup$
My apologies! I forgot one crucial assumption: colinearity of the circle centers!
$endgroup$
– user2471
Mar 21 at 22:28
add a comment |
$begingroup$
I will assume that points $O,O_1,O_2$ are collinear in accordance with your drawing. Otherwise the claim is in general not valid.
Let the points of intersecttion of circle $mathcalC$ with circles $mathcalC_1$ and $mathcalC_2$ be $I_1$ and $I_2$, respectively. Let $angle O_1IA=angle O_2IB$ be $alpha$. Let $R_1$ and $R_2$ be the radii of $mathcalC_1$ and $mathcalC_2$, respectively.
We have
$$
IO=R_2-R_1,quad IB=2R_1cosalpha,quad IC=2R_2cosalpha.
$$
The last two equalities follow from considering the right triangles $IBI_1$ and $ICI_2$.
By the law of cosines one then obtains:
$$beginalign
OB^2&=(R_2-R_1)^2+(2R_1cosalpha)^2+2(R_2-R_1)(2R_1cosalpha)cosalpha
=IO^2+IBcdot IC,\
OC^2&=(R_2-R_1)^2+(2R_2cosalpha)^2-2(R_2-R_1)(2R_2cosalpha)cosalpha
=IO^2+IBcdot IC.
endalign
$$
Thus, $OB=OC$. The rest is simple.
answered Mar 21 at 21:29
useruser
6,21911031
$endgroup$
$begingroup$
OP requires not to use trigonometry.
$endgroup$
– Matteo
Mar 21 at 22:25
$begingroup$
@Matteo I have noticed it. I however do not know if it refers to the usage of cosine law (or Pythagorean theorem). No other "trigonometric" properties are used. In fact $cosalpha$ can be considered here as an abbreviation for $fracIB2R_1=fracIC2R_2$.
$endgroup$
– user
Mar 21 at 22:47
add a comment |
$begingroup$
I will assume that points $O,O_1,O_2$ are collinear in accordance with your drawing. Otherwise the claim is in general not valid.
Let the points of intersecttion of circle $mathcalC$ with circles $mathcalC_1$ and $mathcalC_2$ be $I_1$ and $I_2$, respectively. Let $angle O_1IA=angle O_2IB$ be $alpha$. Let $R_1$ and $R_2$ be the radii of $mathcalC_1$ and $mathcalC_2$, respectively.
We have
$$
IO=R_2-R_1,quad IB=2R_1cosalpha,quad IC=2R_2cosalpha.
$$
The last two equalities follow from considering the right triangles $IBI_1$ and $ICI_2$.
By the law of cosines one then obtains:
$$beginalign
OB^2&=(R_2-R_1)^2+(2R_1cosalpha)^2+2(R_2-R_1)(2R_1cosalpha)cosalpha
=IO^2+IBcdot IC,\
OC^2&=(R_2-R_1)^2+(2R_2cosalpha)^2-2(R_2-R_1)(2R_2cosalpha)cosalpha
=IO^2+IBcdot IC.
endalign
$$
Thus, $OB=OC$. The rest is simple.
answered Mar 21 at 21:29
useruser
6,21911031
$endgroup$
$begingroup$
OP requires not to use trigonometry.
$endgroup$
– Matteo
Mar 21 at 22:25
$begingroup$
@Matteo I have noticed it. I however do not know if it refers to the usage of cosine law (or Pythagorean theorem). No other "trigonometric" properties are used. In fact $cosalpha$ can be considered here as an abbreviation for $fracIB2R_1=fracIC2R_2$.
$endgroup$
– user
Mar 21 at 22:47
add a comment |
$begingroup$
I will assume that points $O,O_1,O_2$ are collinear in accordance with your drawing. Otherwise the claim is in general not valid.
Let the points of intersecttion of circle $mathcalC$ with circles $mathcalC_1$ and $mathcalC_2$ be $I_1$ and $I_2$, respectively. Let $angle O_1IA=angle O_2IB$ be $alpha$. Let $R_1$ and $R_2$ be the radii of $mathcalC_1$ and $mathcalC_2$, respectively.
We have
$$
IO=R_2-R_1,quad IB=2R_1cosalpha,quad IC=2R_2cosalpha.
$$
The last two equalities follow from considering the right triangles $IBI_1$ and $ICI_2$.
By the law of cosines one then obtains:
$$beginalign
OB^2&=(R_2-R_1)^2+(2R_1cosalpha)^2+2(R_2-R_1)(2R_1cosalpha)cosalpha
=IO^2+IBcdot IC,\
OC^2&=(R_2-R_1)^2+(2R_2cosalpha)^2-2(R_2-R_1)(2R_2cosalpha)cosalpha
=IO^2+IBcdot IC.
endalign
$$
Thus, $OB=OC$. The rest is simple.
answered Mar 21 at 21:29
useruser
6,21911031
$endgroup$
I will assume that points $O,O_1,O_2$ are collinear in accordance with your drawing. Otherwise the claim is in general not valid.
Let the points of intersecttion of circle $mathcalC$ with circles $mathcalC_1$ and $mathcalC_2$ be $I_1$ and $I_2$, respectively. Let $angle O_1IA=angle O_2IB$ be $alpha$. Let $R_1$ and $R_2$ be the radii of $mathcalC_1$ and $mathcalC_2$, respectively.
We have
$$
IO=R_2-R_1,quad IB=2R_1cosalpha,quad IC=2R_2cosalpha.
$$
The last two equalities follow from considering the right triangles $IBI_1$ and $ICI_2$.
By the law of cosines one then obtains:
$$beginalign
OB^2&=(R_2-R_1)^2+(2R_1cosalpha)^2+2(R_2-R_1)(2R_1cosalpha)cosalpha
=IO^2+IBcdot IC,\
OC^2&=(R_2-R_1)^2+(2R_2cosalpha)^2-2(R_2-R_1)(2R_2cosalpha)cosalpha
=IO^2+IBcdot IC.
endalign
$$
Thus, $OB=OC$. The rest is simple.
answered Mar 21 at 21:29
useruser
6,21911031
edited Mar 21 at 23:00
answered Mar 21 at 21:29
useruser
6,21911031
answered Mar 21 at 21:29
useruser
6,21911031
answered Mar 21 at 21:29
useruser
6,21911031
6,21911031
$begingroup$
OP requires not to use trigonometry.
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– Matteo
Mar 21 at 22:25
$begingroup$
@Matteo I have noticed it. I however do not know if it refers to the usage of cosine law (or Pythagorean theorem). No other "trigonometric" properties are used. In fact $cosalpha$ can be considered here as an abbreviation for $fracIB2R_1=fracIC2R_2$.
$endgroup$
– user
Mar 21 at 22:47
add a comment |
$begingroup$
OP requires not to use trigonometry.
$endgroup$
– Matteo
Mar 21 at 22:25
$begingroup$
@Matteo I have noticed it. I however do not know if it refers to the usage of cosine law (or Pythagorean theorem). No other "trigonometric" properties are used. In fact $cosalpha$ can be considered here as an abbreviation for $fracIB2R_1=fracIC2R_2$.
$endgroup$
– user
Mar 21 at 22:47
$begingroup$
OP requires not to use trigonometry.
$endgroup$
– Matteo
Mar 21 at 22:25
$begingroup$
OP requires not to use trigonometry.
$endgroup$
– Matteo
Mar 21 at 22:25
$begingroup$
@Matteo I have noticed it. I however do not know if it refers to the usage of cosine law (or Pythagorean theorem). No other "trigonometric" properties are used. In fact $cosalpha$ can be considered here as an abbreviation for $fracIB2R_1=fracIC2R_2$.
$endgroup$
– user
Mar 21 at 22:47
$begingroup$
@Matteo I have noticed it. I however do not know if it refers to the usage of cosine law (or Pythagorean theorem). No other "trigonometric" properties are used. In fact $cosalpha$ can be considered here as an abbreviation for $fracIB2R_1=fracIC2R_2$.
$endgroup$
– user
Mar 21 at 22:47
add a comment |
$begingroup$
Here is a possibile path.
$angle O_1BI congangle O_2CI$ (can you tell why?). Therefore $O_1Bparallel O_2C$.- Consequently $angle BO_1O cong angle CO_2O$.
$triangle O_1BO cong triangle O_2CO$ (can you tell why?).- In particular, $OBcong OC$. Thus $triangle OBC$ is isosceles.
And the thesis follows from what you already noticed, since now you can demonstrate that $triangle ABO cong triangle CDO$, as you correctly wished to show.
answered Mar 21 at 23:01
MatteoMatteo
1,302313
$endgroup$
add a comment |
$begingroup$
Here is a possibile path.
$angle O_1BI congangle O_2CI$ (can you tell why?). Therefore $O_1Bparallel O_2C$.- Consequently $angle BO_1O cong angle CO_2O$.
$triangle O_1BO cong triangle O_2CO$ (can you tell why?).- In particular, $OBcong OC$. Thus $triangle OBC$ is isosceles.
And the thesis follows from what you already noticed, since now you can demonstrate that $triangle ABO cong triangle CDO$, as you correctly wished to show.
answered Mar 21 at 23:01
MatteoMatteo
1,302313
$endgroup$
add a comment |
$begingroup$
Here is a possibile path.
$angle O_1BI congangle O_2CI$ (can you tell why?). Therefore $O_1Bparallel O_2C$.- Consequently $angle BO_1O cong angle CO_2O$.
$triangle O_1BO cong triangle O_2CO$ (can you tell why?).- In particular, $OBcong OC$. Thus $triangle OBC$ is isosceles.
And the thesis follows from what you already noticed, since now you can demonstrate that $triangle ABO cong triangle CDO$, as you correctly wished to show.
answered Mar 21 at 23:01
MatteoMatteo
1,302313
$endgroup$
Here is a possibile path.
$angle O_1BI congangle O_2CI$ (can you tell why?). Therefore $O_1Bparallel O_2C$.- Consequently $angle BO_1O cong angle CO_2O$.
$triangle O_1BO cong triangle O_2CO$ (can you tell why?).- In particular, $OBcong OC$. Thus $triangle OBC$ is isosceles.
And the thesis follows from what you already noticed, since now you can demonstrate that $triangle ABO cong triangle CDO$, as you correctly wished to show.
answered Mar 21 at 23:01
MatteoMatteo
1,302313
edited Mar 21 at 23:13
answered Mar 21 at 23:01
MatteoMatteo
1,302313
answered Mar 21 at 23:01
MatteoMatteo
1,302313
answered Mar 21 at 23:01
MatteoMatteo
1,302313
1,302313
add a comment |
add a comment |
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$begingroup$
In your figure $O$ lies on line $O_1O_2$. Is this an additional hypothesis?
$endgroup$
– Matteo
Mar 21 at 22:07
1
$begingroup$
Yes! I forgot that
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– user2471
Mar 21 at 22:21
$begingroup$
that explains the counter example below...
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– Matteo
Mar 21 at 22:22
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@Matteo my Apologies!
$endgroup$
– user2471
Mar 21 at 22:26
$begingroup$
If you name the endpoints of the green diameter say M and N, then $angle MBI$ and $angle NCI$ are both right, hence right triangles $triangle MBI$ and $triangle NCI$ are similar (with the scale equal to the ratio of radii of small circles); then $I$ divides $BC$ in the same proportion as it divides the green diameter $MN$. Alas, I can't see yet how it can help...
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– CiaPan
Mar 21 at 22:52