Inscribing a rectangle in a triangle [closed]
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How can we inscribe a rectangle of given diagonal in a scalene triangle? How can it be inscribed if its diagonal is the smallest possible?
NOTE : The triangle is not a right angled triangle.
geometry triangle
edited Mar 16 at 20:30
Akari
694119
asked Mar 16 at 15:01
Mayank JangidMayank Jangid
11
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closed as off-topic by Omega Krypton, Rupert Morrish, Peregrine Rook, Rand al'Thor, JonMark Perry Mar 17 at 8:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Omega Krypton, Rupert Morrish, Peregrine Rook, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How can we inscribe a rectangle of given diagonal in a scalene triangle? How can it be inscribed if its diagonal is the smallest possible?
NOTE : The triangle is not a right angled triangle.
geometry triangle
edited Mar 16 at 20:30
Akari
694119
asked Mar 16 at 15:01
Mayank JangidMayank Jangid
11
$endgroup$
closed as off-topic by Omega Krypton, Rupert Morrish, Peregrine Rook, Rand al'Thor, JonMark Perry Mar 17 at 8:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Omega Krypton, Rupert Morrish, Peregrine Rook, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How can we inscribe a rectangle of given diagonal in a scalene triangle? How can it be inscribed if its diagonal is the smallest possible?
NOTE : The triangle is not a right angled triangle.
geometry triangle
edited Mar 16 at 20:30
Akari
694119
asked Mar 16 at 15:01
Mayank JangidMayank Jangid
11
$endgroup$
How can we inscribe a rectangle of given diagonal in a scalene triangle? How can it be inscribed if its diagonal is the smallest possible?
NOTE : The triangle is not a right angled triangle.
geometry triangle
geometry triangle
edited Mar 16 at 20:30
Akari
694119
asked Mar 16 at 15:01
Mayank JangidMayank Jangid
11
edited Mar 16 at 20:30
Akari
694119
asked Mar 16 at 15:01
Mayank JangidMayank Jangid
11
edited Mar 16 at 20:30
Akari
694119
edited Mar 16 at 20:30
Akari
694119
edited Mar 16 at 20:30
Akari
694119
694119
asked Mar 16 at 15:01
Mayank JangidMayank Jangid
11
asked Mar 16 at 15:01
Mayank JangidMayank Jangid
11
asked Mar 16 at 15:01
Mayank JangidMayank Jangid
11
11
closed as off-topic by Omega Krypton, Rupert Morrish, Peregrine Rook, Rand al'Thor, JonMark Perry Mar 17 at 8:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Omega Krypton, Rupert Morrish, Peregrine Rook, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Omega Krypton, Rupert Morrish, Peregrine Rook, Rand al'Thor, JonMark Perry Mar 17 at 8:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Omega Krypton, Rupert Morrish, Peregrine Rook, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a solution:
Let AB be the base of the given triangle ABC. Construct a right triangle ADB so that CD || AB and AD is perpendicular to AB. (Point D is an intersection of a perpendicular to AB and a line drawn through C parallel to AB). Construct AG so that G is on the side DB and AG is equal to the given diagonal of the rectangle. Draw line GE parallel to AB. The line intersects AD at E, AC at F, CB at H. Drop perpendiculars FI, GJ, HK to AB. Rectangle AEGJ has a given diagonal. Rectangle IFHK is the solution because it is equal to the rectangle AEGJ. We can get the smallest possible diagonal when we make AG to be the altitude of the triangle ADB. Here is a diagram:
To prove that rectangle IFHK is equal to rectangle AEGJ, first observe that EA = FI as opposite sides of rectangle AEFI. Second, we will prove EG = FH.
Triangle EDG is similar to triangle ADB (because EG || AB).
Hence EG/AB = DE/DA.
Triangle FCH is similar to triangle ACB (because FH || AB).
Hence FH/AB = CH/CB.
But DE/DA = CH/CB because three parallel lines (DC, EH, AB) divide two transversals (DA and CB) proportionally.
Hence EG/AB = FH/AB and so EG = FH. QED.
answered Mar 16 at 17:12
ppgdevppgdev
19816
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Can you please give the construction's.
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– Mayank Jangid
Mar 17 at 4:04
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Full justification .
$endgroup$
– Mayank Jangid
Mar 17 at 4:05
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Mayank Jangid, could you be more specific about what construction step needs to be clarified. Do you need an explanation why rectangle IFHK is equal to rectangle AEGJ?
$endgroup$
– ppgdev
Mar 17 at 12:06
$begingroup$
Yes, I want to know why rectangle IFHK is equal to rectangle AEGJ.
$endgroup$
– Mayank Jangid
Mar 17 at 13:11
$begingroup$
Mayank Jangid, I added a proof that rectangle IFHK is equal to rectangle AEGJ to my answer. Please, let me know if it answers your question.
$endgroup$
– ppgdev
Mar 17 at 14:34
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a solution:
Let AB be the base of the given triangle ABC. Construct a right triangle ADB so that CD || AB and AD is perpendicular to AB. (Point D is an intersection of a perpendicular to AB and a line drawn through C parallel to AB). Construct AG so that G is on the side DB and AG is equal to the given diagonal of the rectangle. Draw line GE parallel to AB. The line intersects AD at E, AC at F, CB at H. Drop perpendiculars FI, GJ, HK to AB. Rectangle AEGJ has a given diagonal. Rectangle IFHK is the solution because it is equal to the rectangle AEGJ. We can get the smallest possible diagonal when we make AG to be the altitude of the triangle ADB. Here is a diagram:
To prove that rectangle IFHK is equal to rectangle AEGJ, first observe that EA = FI as opposite sides of rectangle AEFI. Second, we will prove EG = FH.
Triangle EDG is similar to triangle ADB (because EG || AB).
Hence EG/AB = DE/DA.
Triangle FCH is similar to triangle ACB (because FH || AB).
Hence FH/AB = CH/CB.
But DE/DA = CH/CB because three parallel lines (DC, EH, AB) divide two transversals (DA and CB) proportionally.
Hence EG/AB = FH/AB and so EG = FH. QED.
answered Mar 16 at 17:12
ppgdevppgdev
19816
$endgroup$
$begingroup$
Can you please give the construction's.
$endgroup$
– Mayank Jangid
Mar 17 at 4:04
$begingroup$
Full justification .
$endgroup$
– Mayank Jangid
Mar 17 at 4:05
$begingroup$
Mayank Jangid, could you be more specific about what construction step needs to be clarified. Do you need an explanation why rectangle IFHK is equal to rectangle AEGJ?
$endgroup$
– ppgdev
Mar 17 at 12:06
$begingroup$
Yes, I want to know why rectangle IFHK is equal to rectangle AEGJ.
$endgroup$
– Mayank Jangid
Mar 17 at 13:11
$begingroup$
Mayank Jangid, I added a proof that rectangle IFHK is equal to rectangle AEGJ to my answer. Please, let me know if it answers your question.
$endgroup$
– ppgdev
Mar 17 at 14:34
|
show 2 more comments
$begingroup$
Here is a solution:
Let AB be the base of the given triangle ABC. Construct a right triangle ADB so that CD || AB and AD is perpendicular to AB. (Point D is an intersection of a perpendicular to AB and a line drawn through C parallel to AB). Construct AG so that G is on the side DB and AG is equal to the given diagonal of the rectangle. Draw line GE parallel to AB. The line intersects AD at E, AC at F, CB at H. Drop perpendiculars FI, GJ, HK to AB. Rectangle AEGJ has a given diagonal. Rectangle IFHK is the solution because it is equal to the rectangle AEGJ. We can get the smallest possible diagonal when we make AG to be the altitude of the triangle ADB. Here is a diagram:
To prove that rectangle IFHK is equal to rectangle AEGJ, first observe that EA = FI as opposite sides of rectangle AEFI. Second, we will prove EG = FH.
Triangle EDG is similar to triangle ADB (because EG || AB).
Hence EG/AB = DE/DA.
Triangle FCH is similar to triangle ACB (because FH || AB).
Hence FH/AB = CH/CB.
But DE/DA = CH/CB because three parallel lines (DC, EH, AB) divide two transversals (DA and CB) proportionally.
Hence EG/AB = FH/AB and so EG = FH. QED.
answered Mar 16 at 17:12
ppgdevppgdev
19816
$endgroup$
$begingroup$
Can you please give the construction's.
$endgroup$
– Mayank Jangid
Mar 17 at 4:04
$begingroup$
Full justification .
$endgroup$
– Mayank Jangid
Mar 17 at 4:05
$begingroup$
Mayank Jangid, could you be more specific about what construction step needs to be clarified. Do you need an explanation why rectangle IFHK is equal to rectangle AEGJ?
$endgroup$
– ppgdev
Mar 17 at 12:06
$begingroup$
Yes, I want to know why rectangle IFHK is equal to rectangle AEGJ.
$endgroup$
– Mayank Jangid
Mar 17 at 13:11
$begingroup$
Mayank Jangid, I added a proof that rectangle IFHK is equal to rectangle AEGJ to my answer. Please, let me know if it answers your question.
$endgroup$
– ppgdev
Mar 17 at 14:34
|
show 2 more comments
$begingroup$
Here is a solution:
Let AB be the base of the given triangle ABC. Construct a right triangle ADB so that CD || AB and AD is perpendicular to AB. (Point D is an intersection of a perpendicular to AB and a line drawn through C parallel to AB). Construct AG so that G is on the side DB and AG is equal to the given diagonal of the rectangle. Draw line GE parallel to AB. The line intersects AD at E, AC at F, CB at H. Drop perpendiculars FI, GJ, HK to AB. Rectangle AEGJ has a given diagonal. Rectangle IFHK is the solution because it is equal to the rectangle AEGJ. We can get the smallest possible diagonal when we make AG to be the altitude of the triangle ADB. Here is a diagram:
To prove that rectangle IFHK is equal to rectangle AEGJ, first observe that EA = FI as opposite sides of rectangle AEFI. Second, we will prove EG = FH.
Triangle EDG is similar to triangle ADB (because EG || AB).
Hence EG/AB = DE/DA.
Triangle FCH is similar to triangle ACB (because FH || AB).
Hence FH/AB = CH/CB.
But DE/DA = CH/CB because three parallel lines (DC, EH, AB) divide two transversals (DA and CB) proportionally.
Hence EG/AB = FH/AB and so EG = FH. QED.
answered Mar 16 at 17:12
ppgdevppgdev
19816
$endgroup$
Here is a solution:
Let AB be the base of the given triangle ABC. Construct a right triangle ADB so that CD || AB and AD is perpendicular to AB. (Point D is an intersection of a perpendicular to AB and a line drawn through C parallel to AB). Construct AG so that G is on the side DB and AG is equal to the given diagonal of the rectangle. Draw line GE parallel to AB. The line intersects AD at E, AC at F, CB at H. Drop perpendiculars FI, GJ, HK to AB. Rectangle AEGJ has a given diagonal. Rectangle IFHK is the solution because it is equal to the rectangle AEGJ. We can get the smallest possible diagonal when we make AG to be the altitude of the triangle ADB. Here is a diagram:
To prove that rectangle IFHK is equal to rectangle AEGJ, first observe that EA = FI as opposite sides of rectangle AEFI. Second, we will prove EG = FH.
Triangle EDG is similar to triangle ADB (because EG || AB).
Hence EG/AB = DE/DA.
Triangle FCH is similar to triangle ACB (because FH || AB).
Hence FH/AB = CH/CB.
But DE/DA = CH/CB because three parallel lines (DC, EH, AB) divide two transversals (DA and CB) proportionally.
Hence EG/AB = FH/AB and so EG = FH. QED.
answered Mar 16 at 17:12
ppgdevppgdev
19816
edited Mar 17 at 14:31
answered Mar 16 at 17:12
ppgdevppgdev
19816
answered Mar 16 at 17:12
ppgdevppgdev
19816
answered Mar 16 at 17:12
ppgdevppgdev
19816
19816
$begingroup$
Can you please give the construction's.
$endgroup$
– Mayank Jangid
Mar 17 at 4:04
$begingroup$
Full justification .
$endgroup$
– Mayank Jangid
Mar 17 at 4:05
$begingroup$
Mayank Jangid, could you be more specific about what construction step needs to be clarified. Do you need an explanation why rectangle IFHK is equal to rectangle AEGJ?
$endgroup$
– ppgdev
Mar 17 at 12:06
$begingroup$
Yes, I want to know why rectangle IFHK is equal to rectangle AEGJ.
$endgroup$
– Mayank Jangid
Mar 17 at 13:11
$begingroup$
Mayank Jangid, I added a proof that rectangle IFHK is equal to rectangle AEGJ to my answer. Please, let me know if it answers your question.
$endgroup$
– ppgdev
Mar 17 at 14:34
|
show 2 more comments
$begingroup$
Can you please give the construction's.
$endgroup$
– Mayank Jangid
Mar 17 at 4:04
$begingroup$
Full justification .
$endgroup$
– Mayank Jangid
Mar 17 at 4:05
$begingroup$
Mayank Jangid, could you be more specific about what construction step needs to be clarified. Do you need an explanation why rectangle IFHK is equal to rectangle AEGJ?
$endgroup$
– ppgdev
Mar 17 at 12:06
$begingroup$
Yes, I want to know why rectangle IFHK is equal to rectangle AEGJ.
$endgroup$
– Mayank Jangid
Mar 17 at 13:11
$begingroup$
Mayank Jangid, I added a proof that rectangle IFHK is equal to rectangle AEGJ to my answer. Please, let me know if it answers your question.
$endgroup$
– ppgdev
Mar 17 at 14:34
$begingroup$
Can you please give the construction's.
$endgroup$
– Mayank Jangid
Mar 17 at 4:04
$begingroup$
Can you please give the construction's.
$endgroup$
– Mayank Jangid
Mar 17 at 4:04
$begingroup$
Full justification .
$endgroup$
– Mayank Jangid
Mar 17 at 4:05
$begingroup$
Full justification .
$endgroup$
– Mayank Jangid
Mar 17 at 4:05
$begingroup$
Mayank Jangid, could you be more specific about what construction step needs to be clarified. Do you need an explanation why rectangle IFHK is equal to rectangle AEGJ?
$endgroup$
– ppgdev
Mar 17 at 12:06
$begingroup$
Mayank Jangid, could you be more specific about what construction step needs to be clarified. Do you need an explanation why rectangle IFHK is equal to rectangle AEGJ?
$endgroup$
– ppgdev
Mar 17 at 12:06
$begingroup$
Yes, I want to know why rectangle IFHK is equal to rectangle AEGJ.
$endgroup$
– Mayank Jangid
Mar 17 at 13:11
$begingroup$
Yes, I want to know why rectangle IFHK is equal to rectangle AEGJ.
$endgroup$
– Mayank Jangid
Mar 17 at 13:11
$begingroup$
Mayank Jangid, I added a proof that rectangle IFHK is equal to rectangle AEGJ to my answer. Please, let me know if it answers your question.
$endgroup$
– ppgdev
Mar 17 at 14:34
$begingroup$
Mayank Jangid, I added a proof that rectangle IFHK is equal to rectangle AEGJ to my answer. Please, let me know if it answers your question.
$endgroup$
– ppgdev
Mar 17 at 14:34
|
show 2 more comments
