Does the quadrilateral have an inscribed circle?
Question: Let o1, o2 be circles inside an angle tangent to one of its sides in
points A, B and to the other in points C, D. Prove that if o1, o2 are externally
tangent, then ABCD has an inscribed circle.
What I have so far:
1.|AB| = |CD| by the strongest theorem of geometry
- A circle can be inscribed in a quadrilateral if and only if the addition of its opposite sides are equal ie. |AB|+|CD|=|BC|+|AD| (depends on how you label the vertices).
So we want to prove that 2|AB|=|BC|+|AD| and it must have to do with the circles being externally tangent.
I do not know how to proceed. Any help is much appreciated.
geometry euclidean-geometry
asked Dec 12 '18 at 8:26
ricorico
806
add a comment |
Question: Let o1, o2 be circles inside an angle tangent to one of its sides in
points A, B and to the other in points C, D. Prove that if o1, o2 are externally
tangent, then ABCD has an inscribed circle.
What I have so far:
1.|AB| = |CD| by the strongest theorem of geometry
- A circle can be inscribed in a quadrilateral if and only if the addition of its opposite sides are equal ie. |AB|+|CD|=|BC|+|AD| (depends on how you label the vertices).
So we want to prove that 2|AB|=|BC|+|AD| and it must have to do with the circles being externally tangent.
I do not know how to proceed. Any help is much appreciated.
geometry euclidean-geometry
asked Dec 12 '18 at 8:26
ricorico
806
I'm not sure you have chosen the best result from which to start. There is some symmetry in the situation which may help.
– Mark Bennet
Dec 12 '18 at 8:50
Please add a diagram.
– Anubhab Ghosal
Dec 12 '18 at 11:10
add a comment |
Question: Let o1, o2 be circles inside an angle tangent to one of its sides in
points A, B and to the other in points C, D. Prove that if o1, o2 are externally
tangent, then ABCD has an inscribed circle.
What I have so far:
1.|AB| = |CD| by the strongest theorem of geometry
- A circle can be inscribed in a quadrilateral if and only if the addition of its opposite sides are equal ie. |AB|+|CD|=|BC|+|AD| (depends on how you label the vertices).
So we want to prove that 2|AB|=|BC|+|AD| and it must have to do with the circles being externally tangent.
I do not know how to proceed. Any help is much appreciated.
geometry euclidean-geometry
asked Dec 12 '18 at 8:26
ricorico
806
Question: Let o1, o2 be circles inside an angle tangent to one of its sides in
points A, B and to the other in points C, D. Prove that if o1, o2 are externally
tangent, then ABCD has an inscribed circle.
What I have so far:
1.|AB| = |CD| by the strongest theorem of geometry
- A circle can be inscribed in a quadrilateral if and only if the addition of its opposite sides are equal ie. |AB|+|CD|=|BC|+|AD| (depends on how you label the vertices).
So we want to prove that 2|AB|=|BC|+|AD| and it must have to do with the circles being externally tangent.
I do not know how to proceed. Any help is much appreciated.
geometry euclidean-geometry
geometry euclidean-geometry
asked Dec 12 '18 at 8:26
ricorico
806
asked Dec 12 '18 at 8:26
ricorico
806
asked Dec 12 '18 at 8:26
ricorico
806
asked Dec 12 '18 at 8:26
ricorico
806
asked Dec 12 '18 at 8:26
ricorico
806
806
I'm not sure you have chosen the best result from which to start. There is some symmetry in the situation which may help.
– Mark Bennet
Dec 12 '18 at 8:50
Please add a diagram.
– Anubhab Ghosal
Dec 12 '18 at 11:10
add a comment |
I'm not sure you have chosen the best result from which to start. There is some symmetry in the situation which may help.
– Mark Bennet
Dec 12 '18 at 8:50
Please add a diagram.
– Anubhab Ghosal
Dec 12 '18 at 11:10
I'm not sure you have chosen the best result from which to start. There is some symmetry in the situation which may help.
– Mark Bennet
Dec 12 '18 at 8:50
I'm not sure you have chosen the best result from which to start. There is some symmetry in the situation which may help.
– Mark Bennet
Dec 12 '18 at 8:50
Please add a diagram.
– Anubhab Ghosal
Dec 12 '18 at 11:10
Please add a diagram.
– Anubhab Ghosal
Dec 12 '18 at 11:10
add a comment |
2 Answers
2
active
oldest
votes
From the diagram, $AB=CD=(r_1+r_2)cosalpha$.
$AC=2r_1cosalpha$ and $BD=2r_2cosalpha$.
Therefore, $AB+CD=AC+BD$, and $ABDC$ is a tangential quadrilateral.
$blacksquare$
answered Dec 12 '18 at 11:59
Anubhab GhosalAnubhab Ghosal
81617
add a comment |
We can go a little further and show that the common tangent point of the two circles is the center of the inscribed circle of $ABCD$. In deed, by symmetry, we have $stackrel{frown}{BE} = stackrel{frown}{CE}$. Since $AB$ is tangent to $(O_2)$, it follows that $$angle ABE = frac12stackrel{frown}{BE} = frac12stackrel{frown}{CE} = angle EBC.$$
So $E$ lies on the angle bisector of $angle ABC$. Similarly for all the other angles.
answered Dec 12 '18 at 14:37
Quang HoangQuang Hoang
12.8k1131
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
From the diagram, $AB=CD=(r_1+r_2)cosalpha$.
$AC=2r_1cosalpha$ and $BD=2r_2cosalpha$.
Therefore, $AB+CD=AC+BD$, and $ABDC$ is a tangential quadrilateral.
$blacksquare$
answered Dec 12 '18 at 11:59
Anubhab GhosalAnubhab Ghosal
81617
add a comment |
From the diagram, $AB=CD=(r_1+r_2)cosalpha$.
$AC=2r_1cosalpha$ and $BD=2r_2cosalpha$.
Therefore, $AB+CD=AC+BD$, and $ABDC$ is a tangential quadrilateral.
$blacksquare$
answered Dec 12 '18 at 11:59
Anubhab GhosalAnubhab Ghosal
81617
add a comment |
From the diagram, $AB=CD=(r_1+r_2)cosalpha$.
$AC=2r_1cosalpha$ and $BD=2r_2cosalpha$.
Therefore, $AB+CD=AC+BD$, and $ABDC$ is a tangential quadrilateral.
$blacksquare$
answered Dec 12 '18 at 11:59
Anubhab GhosalAnubhab Ghosal
81617
From the diagram, $AB=CD=(r_1+r_2)cosalpha$.
$AC=2r_1cosalpha$ and $BD=2r_2cosalpha$.
Therefore, $AB+CD=AC+BD$, and $ABDC$ is a tangential quadrilateral.
$blacksquare$
answered Dec 12 '18 at 11:59
Anubhab GhosalAnubhab Ghosal
81617
edited Dec 31 '18 at 7:57
answered Dec 12 '18 at 11:59
Anubhab GhosalAnubhab Ghosal
81617
answered Dec 12 '18 at 11:59
Anubhab GhosalAnubhab Ghosal
81617
answered Dec 12 '18 at 11:59
Anubhab GhosalAnubhab Ghosal
81617
81617
add a comment |
add a comment |
We can go a little further and show that the common tangent point of the two circles is the center of the inscribed circle of $ABCD$. In deed, by symmetry, we have $stackrel{frown}{BE} = stackrel{frown}{CE}$. Since $AB$ is tangent to $(O_2)$, it follows that $$angle ABE = frac12stackrel{frown}{BE} = frac12stackrel{frown}{CE} = angle EBC.$$
So $E$ lies on the angle bisector of $angle ABC$. Similarly for all the other angles.
answered Dec 12 '18 at 14:37
Quang HoangQuang Hoang
12.8k1131
add a comment |
We can go a little further and show that the common tangent point of the two circles is the center of the inscribed circle of $ABCD$. In deed, by symmetry, we have $stackrel{frown}{BE} = stackrel{frown}{CE}$. Since $AB$ is tangent to $(O_2)$, it follows that $$angle ABE = frac12stackrel{frown}{BE} = frac12stackrel{frown}{CE} = angle EBC.$$
So $E$ lies on the angle bisector of $angle ABC$. Similarly for all the other angles.
answered Dec 12 '18 at 14:37
Quang HoangQuang Hoang
12.8k1131
add a comment |
We can go a little further and show that the common tangent point of the two circles is the center of the inscribed circle of $ABCD$. In deed, by symmetry, we have $stackrel{frown}{BE} = stackrel{frown}{CE}$. Since $AB$ is tangent to $(O_2)$, it follows that $$angle ABE = frac12stackrel{frown}{BE} = frac12stackrel{frown}{CE} = angle EBC.$$
So $E$ lies on the angle bisector of $angle ABC$. Similarly for all the other angles.
answered Dec 12 '18 at 14:37
Quang HoangQuang Hoang
12.8k1131
We can go a little further and show that the common tangent point of the two circles is the center of the inscribed circle of $ABCD$. In deed, by symmetry, we have $stackrel{frown}{BE} = stackrel{frown}{CE}$. Since $AB$ is tangent to $(O_2)$, it follows that $$angle ABE = frac12stackrel{frown}{BE} = frac12stackrel{frown}{CE} = angle EBC.$$
So $E$ lies on the angle bisector of $angle ABC$. Similarly for all the other angles.
answered Dec 12 '18 at 14:37
Quang HoangQuang Hoang
12.8k1131
answered Dec 12 '18 at 14:37
Quang HoangQuang Hoang
12.8k1131
answered Dec 12 '18 at 14:37
Quang HoangQuang Hoang
12.8k1131
answered Dec 12 '18 at 14:37
Quang HoangQuang Hoang
12.8k1131
12.8k1131
add a comment |
add a comment |
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I'm not sure you have chosen the best result from which to start. There is some symmetry in the situation which may help.
– Mark Bennet
Dec 12 '18 at 8:50
Please add a diagram.
– Anubhab Ghosal
Dec 12 '18 at 11:10