## InterA.TriangleMedianRightAngle History

November 25, 2009, at 01:39 AM by LFS -
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[]Suitable for 6th-grade on up. Can be connected to Thale's Theorem on Circles, Diameters and Right-Triangles where the "longest side" is the diameter and the median is any radius.

November 25, 2009, at 01:17 AM by LFS -
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• [[GlossaryT/ThalesTheorem|Thales Theorem - Circles, Diameters and Right Triangles]
to:
November 25, 2009, at 01:17 AM by LFS -
November 25, 2009, at 12:39 AM by LFS -
• [[GlossaryT/ThalesTheorem|Thales Theorem - Circles, Diameters and Right Triangles]
January 28, 2009, at 12:26 PM by LFS -
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November 12, 2008, at 04:03 PM by Jeanette Shannon - Comment added
September 23, 2008, at 07:39 AM by LFS -
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[c]

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pdf Δ  or  doc Δ

to:

pdf  or  doc

March 14, 2008, at 03:49 PM by LFS -
March 14, 2008, at 02:32 PM by LFS -
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[]User interacts with construction and notes that the triangle "appears" to be a right triangle. Then, using only the fact that the lower angles of an isosceles triangle are equal, the user "proves" that the angle is in fact 90°. Finally, user constructs the triangle.

to:

[]User interacts with construction and notices that the triangle constructed under the given conditions "appears" to be a right triangle. He then uses standard 6th grade geometric facts to "prove" that the angle is in fact 90°. Finally, in a second interactivity the user constructs the triangle.

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[]Understanding triangles and logical conclusions.

to:

[]Understanding triangles, angles and logical conclusions.

March 12, 2008, at 09:29 AM by LFS -
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[row]

[]
Online Activities
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Online Activity

[tableend]

March 12, 2008, at 09:26 AM by LFS -

[tableend]

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March 12, 2008, at 09:08 AM by LFS -

[ valign=middle]Pre-knowledge
[]The sum of the angles of a triangle is 180°, the sum of supplementary angles is 180° the angles at the base of an isosceles triangle are congruent (equal size).
[row]

February 07, 2008, at 02:52 PM by LFS -
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February 07, 2008, at 02:49 PM by LFS -
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February 07, 2008, at 02:48 PM by LFS -
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February 07, 2008, at 02:37 PM by LFS -
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Finally, why does angle < ACB = 90^\circ ?
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Finally, why does < ACB = 90^\circ ?
February 07, 2008, at 04:11 AM by LFS -
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(You can also construct \overline{CA} "by hand" with a circle with center C and radius "a" and point A - any point on circle.)
to:
(You can also construct \overline{CA} "by hand" with a circle with center C and radius "a" and point A - any point on this circle.)
February 07, 2008, at 04:10 AM by LFS -
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(You can also construct \overline{CA} "by hand" with a ray from C (rename the second point P), a circle with center C and radius "a" and then A is the intersection point.)
to:
(You can also construct \overline{CA} "by hand" with a circle with center C and radius "a" and point A - any point on circle.)
February 07, 2008, at 12:27 AM by LFS -
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1. What kind of triangle is \Delta BMC ?
2. Why does \frac{\alpha}{2}+ \gamma = 90^\circ ?
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February 06, 2008, at 11:07 PM by LFS -
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February 06, 2008, at 11:00 PM by LFS -
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• You need to construct the solid lines!  To construct \overline{CA}
• Click on Attach:GgbActivity/segment_len.jpg Δ, then on point C and then type in "a" (letter a - no quotes) and hit Enter.
to:
• You need to construct the solid lines!
• To construct \overline{CA} , click on Attach:GgbActivity/segment_len.jpg Δ, then on point C and then type in "a" (letter a - no quotes) and hit Enter.
February 06, 2008, at 10:59 PM by LFS -
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2. Notice that triangle ABC satisfies the theorem no matter slider or point positions. This is because of the construction!

to:

2. Notice that ΔABC satisfies the theorem no matter slider or point positions. This is because of the construction!

• You need to construct the solid lines!  To construct \overline{CA}
(You can also construct \overline{CA} "by hand" with a ray from C (rename the second point P), a circle with center C and radius "a" and then A is the intersection point.)
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• Do the same for point A. Then find the intersection point M ...
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• Do the same for point A. Then find the intersection point and name it M ...
February 06, 2008, at 11:23 AM by LFS -
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February 06, 2008, at 11:20 AM by LFS -
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February 06, 2008, at 04:40 AM by LFS -
February 06, 2008, at 03:18 AM by LFS -
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If \overline{CM}\,\, is exactly half the length \overline{AB}\,\, then \Delta ABC \,\,is a right-triangle.
to:
If \overline{CM}\,\, is exactly half the length \overline{AB}\,\, then \Delta ABC \,\,is a right-triangle.
February 06, 2008, at 02:44 AM by LFS -
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February 06, 2008, at 02:43 AM by LFS -
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• Does triangle ABC satisfy the "hypothesis" of the theorem (no matter slider or point positions)? Why do you think this is so?
to:
• Does triangle ABC satisfy the "hypothesis" of the theorem (no matter slider or point positions)?
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Now, why does angle < ACB = 90^\circ ?
to:
Finally, why does angle < ACB = 90^\circ ?
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1. Look at the construction in the window above .

February 06, 2008, at 02:35 AM by LFS -
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If \overline{CM}\,\, is exactly half the length \overline{AB}\,\, then \Delta ABC \,\,is a right triangle.
to:
If \overline{CM}\,\, is exactly half the length \overline{AB}\,\, then \Delta ABC \,\,is a right-triangle.
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• Click the checkbox to show triangle ABC.
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Notice that angle <ACB \,\,appears to a right-angle, i.e. 90^\circ .
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Notice that angle <ACB \,\,appears to a right-angle (90°).

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• Deselect checkbox to hide triangle ABC
1. How many degrees is: \alpha + \beta  ?
to:
• Select checkbox: Show Angles.
1. How many degrees in: \alpha + \beta  ?
February 05, 2008, at 01:02 PM by LFS -
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• Click and drag point C to move and point A to rotate.
to:
• Click and drag point C to move and point A to rotate.
February 05, 2008, at 12:57 PM by LFS -
Changed line 83 from:
• Click and drag point C to move and point A to rotate.
to:
• Click and drag point C to move and point A to rotate.
February 05, 2008, at 12:55 PM by LFS -
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1. What is: \alpha + \beta  ?
to:
1. How many degrees is: \alpha + \beta  ?
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1. What is: 2 \cdot \delta + \beta ?
to:
1. How many degrees in: 2 \cdot \delta + \beta ?
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1. What kind of triangle is \Delta CBM ?
to:
1. What kind of triangle is \Delta BMC ?
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Why does angle < ACB = 90^\circ
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Deleted lines 109-111:
February 05, 2008, at 12:50 PM by LFS -
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[l]Theorem: Let \overline{CM}\,\, be the median to the longest side of \overline{AB}\,\, of triangle \Delta ABC .

to:

[l]Theorem: Let \overline{CM}\,\, be the median to the longest side   \overline{AB}\,\, of triangle \Delta ABC .

February 05, 2008, at 12:49 PM by LFS -
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If \overline{CM}\,\, is exactly half the length \overline{AB}   then \Delta ABC is a right triangle.

[tableend]

If \overline{CM}\,\, is exactly half the length \overline{AB}\,\, then \Delta ABC \,\,is a right triangle.

[tableend]

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Notice that angle <ACB  appears to a right-angle, i.e. 90^\circ .
to:
Notice that angle <ACB \,\,appears to a right-angle, i.e. 90^\circ .
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What part of the theorem have we shown?
to:
What exactly have we shown?
February 05, 2008, at 12:47 PM by LFS -
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[table border=1 width=600px bgcolor=#ffffee]

to:

[table border=1 width=650px bgcolor=#ffffee]

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[l]Theorem: : Let m be the median to the longest side of \overline{AB} of triangle \Delta ABC .

That is, m is the line segment joining the point C to the midpoint M of \overline{AB} .
If m is exactly half the length \overline{AB}   then \Delta ABC is a right triangle.
to:

[l]Theorem: Let \overline{CM}\,\, be the median to the longest side of \overline{AB}\,\, of triangle \Delta ABC .

If \overline{CM}\,\, is exactly half the length \overline{AB}   then \Delta ABC is a right triangle.
February 01, 2008, at 01:02 PM by LFS -
Changed line 33 from:

[]User interacts with construction and notes that the triangle "appears" to be a right triangle. Then, using only the fact that the lower angles of an isosceles triangle are equal, the user "proves" that the angle is in fact 90°. Finally, the student constructs the same interactivity.

to:

[]User interacts with construction and notes that the triangle "appears" to be a right triangle. Then, using only the fact that the lower angles of an isosceles triangle are equal, the user "proves" that the angle is in fact 90°. Finally, user constructs the triangle.

February 01, 2008, at 12:40 PM by LFS -
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• Do the same for point A. Then find the intersection point M ...
February 01, 2008, at 12:30 PM by LFS -
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pdf Δ  or  doc Δ

to:

pdf Δ  or  doc Δ

February 01, 2008, at 12:30 PM by LFS -
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pdf Δ  or  doc Δ

to:

pdf Δ  or  doc Δ

February 01, 2008, at 12:11 PM by LFS -
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[l]Theorem: : Let M be the median to the longest side of \overline{AB} of triangle \Delta ABC .

That is, M is the line segment joining the point C to the midpoint of \overline{AB} .
If M is exactly half the length \overline{AB}   then \Delta ABC is a right triangle.
to:

[l]Theorem: : Let m be the median to the longest side of \overline{AB} of triangle \Delta ABC .

That is, m is the line segment joining the point C to the midpoint M of \overline{AB} .
If m is exactly half the length \overline{AB}   then \Delta ABC is a right triangle.
February 01, 2008, at 12:04 PM by LFS -
February 01, 2008, at 12:03 PM by LFS -
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Online Activity

February 01, 2008, at 11:41 AM by LFS -
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[]Brief

to:

[ valign=middle]Brief

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[]Requires freeware GeoGebra for offline use.

to:

[]Zip for offline use: tri_median_right.zip (includes handout, teachers page, 2 ggb interactivities)
Requires freeware GeoGebra for offline use.

February 01, 2008, at 11:30 AM by LFS -
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February 01, 2008, at 11:30 AM by LFS -
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February 01, 2008, at 11:29 AM by LFS -
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• Does triangle ABC satisfy the "hypothesis" of the theorem (no matter slider or point positions)? Why do you think this is so?
• Does triangle ABC satisfy the "hypothesis" of the theorem (no matter slider or point positions)? Why do you think this is so?
February 01, 2008, at 11:26 AM by LFS -
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February 01, 2008, at 11:22 AM by LFS -
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[]User interacts with construction and notes that the triangle "appears" to be a right triangle. Then, using only the fact that the lower angles of an isoceles triangle are equal, the user "proves" that the angle is in fact 90°. Finally, the student constructs the same interactivity.

to:

[]User interacts with construction and notes that the triangle "appears" to be a right triangle. Then, using only the fact that the lower angles of an isosceles triangle are equal, the user "proves" that the angle is in fact 90°. Finally, the student constructs the same interactivity.

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[]triangles

to:

[]triangles, medians, right-triangles, construction, interactivity, geogebra

February 01, 2008, at 11:12 AM by LFS -
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[l]Theorem: A triangle is a right triangle if and only if the median to the longest side
(hypotenuse) is exactly half the length of this side.

to:

[l]Theorem: : Let M be the median to the longest side of \overline{AB} of triangle \Delta ABC .

That is, M is the line segment joining the point C to the midpoint of \overline{AB} .
If M is exactly half the length \overline{AB}   then \Delta ABC is a right triangle.
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• Click and drag the point C to move and the point A to rotate.
to:
• Click and drag point C to move and point A to rotate.
• Does triangle ABC satisfy the "hypothesis" of the theorem (no matter slider or point positions)? Why do you think this is so?
February 01, 2008, at 10:03 AM by LFS -
February 01, 2008, at 09:49 AM by LFS -
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February 01, 2008, at 09:41 AM by LFS -
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[]Requires freeware GeoGebra for offline use.

to:

[]Requires freeware GeoGebra for offline use.

February 01, 2008, at 09:13 AM by LFS -
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Why does angle < ACB = 90^\circ
to:
Why does angle < ACB = 90^\circ
February 01, 2008, at 09:12 AM by LFS -
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Notice that &208; ACB  appears to a right-angle, i.e. 90^\circ .
to:
Notice that angle <ACB  appears to a right-angle, i.e. 90^\circ .
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Why does &208; ACB = 90^\circ
to:
Why does angle < ACB = 90^\circ
February 01, 2008, at 09:07 AM by LFS -
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Notice that &208; ACB appears to a right-angle, i.e. 90^\circ .
to:
Notice that &208; ACB  appears to a right-angle, i.e. 90^\circ .
February 01, 2008, at 09:07 AM by LFS -
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Notice that \angle ACB appears to a right-angle, i.e. 90^\circ .
to:
Notice that &208; ACB appears to a right-angle, i.e. 90^\circ .
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1. What is: \alpha + \beta ?
2. What kind of triangle is \triangle CAM
3. Why does \delta =\frac{\alpha}{2} ?
to:
1. What is: \alpha + \beta  ?
2. What kind of triangle is \Delta CAM  ?
3. What is: 2 \cdot \delta + \beta ?
4. Why does \delta =\frac{\alpha}{2} ?
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1. What kind of triangle is \triangle CBM
2. Why does \frac{\alpha}{2}+ \gamma = 90^\circ ?
to:
1. What kind of triangle is \Delta CBM ?
2. Why does \frac{\alpha}{2}+ \gamma = 90^\circ ?
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Why does \angle ACB = 90^\circ
to:
Why does &208; ACB = 90^\circ
February 01, 2008, at 08:54 AM by LFS -
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• Click and drag the slider c' to change the length of the median (and the hypotenuse \overline{AB} ).
to:
• Click and drag the slider m to change the length of the median (and the hypotenuse \overline{AB} ).
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1. What kind of triangle is \triangle CAC'
to:
1. What kind of triangle is \triangle CAM
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1. What kind of triangle is \triangle CBC'
to:
1. What kind of triangle is \triangle CBM
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February 01, 2008, at 08:50 AM by LFS -
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February 01, 2008, at 08:49 AM by LFS -
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February 01, 2008, at 08:48 AM by LFS -
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showhide init=hide div=div9 lshow="+" lhide="-"
)Online Activity Sheet - Meta Data
to:
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[]Geometry and Measurement, Geometry

to:

[]Measurement and Geometry, Geometry

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February 01, 2008, at 08:09 AM by LFS -
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Changed line 3 from:

Online Activity Sheet

to:
showhide init=hide div=div9 lshow="+" lhide="-"
)Online Activity Sheet - Meta Data
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[table border=1 width=600px bgcolor=#ffffee]

to:
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[l]Theorem: A triangle is a right triangle if and only if the median to the longest side
(hypotenuse) is exactly half the length of this side.

to:

[]Brief
[]User interacts with construction and notes that the triangle "appears" to be a right triangle. Then, using only the fact that the lower angles of an isoceles triangle are equal, the user "proves" that the angle is in fact 90°. Finally, the student constructs the same interactivity.
[row]
[]Goal
[]Understanding triangles and logical conclusions.
[row]
[row]
[]Strand
[]Geometry and Measurement, Geometry
[row]
[]Standard
[]CA 6.MG.2.2, CA 7.AF.4.2, IS 1.AL.2.4, ACT EE 24-27, ACT GR 24-27
[row]
[]Keywords
[]triangles
[row]
[row]
[]Source
[row]
[]Cost
[]Activity and software is free to use
[row]
[]Requires freeware GeoGebra for offline use.
[row]
[]Type
[]Java Applet so requires free sunJava player

[table border=1 width=600px bgcolor=#ffffee]
[row]
[l]Theorem: A triangle is a right triangle if and only if the median to the longest side
(hypotenuse) is exactly half the length of this side.
[tableend]

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Notice that \angle ACB appears to be 90^\circ .
to:
Notice that \angle ACB appears to a right-angle, i.e. 90^\circ .
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January 30, 2008, at 01:52 PM by LFS -
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January 30, 2008, at 01:48 PM by LFS -
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Some starting hints or complete directions.
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January 30, 2008, at 01:41 PM by LFS -
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January 30, 2008, at 01:37 PM by LFS -
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[l]Theorem: A triangle is a right triangle if and only if the median to the longest side

to:

[l]Theorem: A triangle is a right triangle if and only if the median to the longest side

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January 30, 2008, at 01:28 PM by LFS -
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January 30, 2008, at 01:25 PM by LFS -
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[table width=100%]

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[l width=50%]1. Look at the construction in the window above..
2. Now create this construction in the window below.

to:
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January 30, 2008, at 01:21 PM by LFS -
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[table border=1 width=600px bgcolor=#ffffee]
[row]
[l]Theorem: A triangle is a right triangle if and only if the median to the longest side
(hypotenuse) is exactly half the length of this side.
[tableend]

Changed lines 16-17 from:
• Click and drag the slider a to change the length of the side \overline{AC}
• Click and drag the slider c' to change the length of the median (and the hypotenuse \overline{AB} ).
to:
• Click and drag the slider a to change the length of the side \overline{AC}
• Click and drag the slider c' to change the length of the median (and the hypotenuse \overline{AB} ).
January 30, 2008, at 01:11 PM by LFS -
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Online Activity Sheet

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1. A triangle is a right-angle triangle if and only if the median to the hypotenuse (longest side) is exactly half the length of this side.

to:

1. A triangle is a right triangle if and only if the median to the hypotenuse (longest side) is exactly half the length of this side.

January 30, 2008, at 01:09 PM by LFS -
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Definition: In geometry, the median of a triangle is a line segment joining a vertex and the midpoint of the opposite side.

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1. A triangle is a right-angle triangle if and only if the median to the hypotenuse (longest side) is exactly half the length of this side.

to:

1. A triangle is a right-angle triangle if and only if the median to the hypotenuse (longest side) is exactly half the length of this side.

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• Click and drag the slider a to change the length of the side \overline{BC}
to:
• Click and drag the slider a to change the length of the side \overline{AC}
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• Click and drag the point C to move and the point B to rotate.
• Click the checkbox to show/hide triangle ABC.
to:
• Click and drag the point C to move and the point A to rotate.
• Click the checkbox to show triangle ABC.
Notice that \angle ACB appears to be 90^\circ .
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