Mixing
Proving a right triangle with an altitude have two similar triangles
$begingroup$
Q. Create a right triangle and draw an altitude to the hypotenuse.
When dragging the points of the right triangle, noticed that the two
smaller triangles that are formed within the larger right triangle
appear to always be similar to each other, and more surprisingly, seem
to always be similar to the big triangle. How do you prove it?
I am having a hard time proving this because I was taught that if we want to prove triangles similar we need two angles but I can only find one.
Let ABC be a right triangle with altitude AD. I need to show triangle ADB and ADC are similar. ADB and ADC are right angles so they are equal to each other.
Any idea how to get the 2nd angle?
geometry
asked Mar 16 '16 at 20:17
JustinJustin
6403916
$endgroup$
add a comment |
$begingroup$
Q. Create a right triangle and draw an altitude to the hypotenuse.
When dragging the points of the right triangle, noticed that the two
smaller triangles that are formed within the larger right triangle
appear to always be similar to each other, and more surprisingly, seem
to always be similar to the big triangle. How do you prove it?
I am having a hard time proving this because I was taught that if we want to prove triangles similar we need two angles but I can only find one.
Let ABC be a right triangle with altitude AD. I need to show triangle ADB and ADC are similar. ADB and ADC are right angles so they are equal to each other.
Any idea how to get the 2nd angle?
geometry
asked Mar 16 '16 at 20:17
JustinJustin
6403916
$endgroup$
1
$begingroup$
It is easiest to show $ABC$ and $ABD$ are similar. They share an angle at $B$.
$endgroup$
– André Nicolas
Mar 16 '16 at 20:21
add a comment |
$begingroup$
Q. Create a right triangle and draw an altitude to the hypotenuse.
When dragging the points of the right triangle, noticed that the two
smaller triangles that are formed within the larger right triangle
appear to always be similar to each other, and more surprisingly, seem
to always be similar to the big triangle. How do you prove it?
I am having a hard time proving this because I was taught that if we want to prove triangles similar we need two angles but I can only find one.
Let ABC be a right triangle with altitude AD. I need to show triangle ADB and ADC are similar. ADB and ADC are right angles so they are equal to each other.
Any idea how to get the 2nd angle?
geometry
asked Mar 16 '16 at 20:17
JustinJustin
6403916
$endgroup$
Q. Create a right triangle and draw an altitude to the hypotenuse.
When dragging the points of the right triangle, noticed that the two
smaller triangles that are formed within the larger right triangle
appear to always be similar to each other, and more surprisingly, seem
to always be similar to the big triangle. How do you prove it?
I am having a hard time proving this because I was taught that if we want to prove triangles similar we need two angles but I can only find one.
Let ABC be a right triangle with altitude AD. I need to show triangle ADB and ADC are similar. ADB and ADC are right angles so they are equal to each other.
Any idea how to get the 2nd angle?
geometry
geometry
asked Mar 16 '16 at 20:17
JustinJustin
6403916
asked Mar 16 '16 at 20:17
JustinJustin
6403916
asked Mar 16 '16 at 20:17
JustinJustin
6403916
asked Mar 16 '16 at 20:17
JustinJustin
6403916
asked Mar 16 '16 at 20:17
JustinJustin
6403916
6403916
1
$begingroup$
It is easiest to show $ABC$ and $ABD$ are similar. They share an angle at $B$.
$endgroup$
– André Nicolas
Mar 16 '16 at 20:21
add a comment |
1
$begingroup$
It is easiest to show $ABC$ and $ABD$ are similar. They share an angle at $B$.
$endgroup$
– André Nicolas
Mar 16 '16 at 20:21
1
1
$begingroup$
It is easiest to show $ABC$ and $ABD$ are similar. They share an angle at $B$.
$endgroup$
– André Nicolas
Mar 16 '16 at 20:21
$begingroup$
It is easiest to show $ABC$ and $ABD$ are similar. They share an angle at $B$.
$endgroup$
– André Nicolas
Mar 16 '16 at 20:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Whether you add angles DAC to BAD or ACD, the result ( sum total) is same, equal to 90 degrees. So they are equal.
$$ angle BAD = angle ACD $$
similarly
$$ angle DBA= angle DAC $$
You said
$$ angle ADB = angle CDA = 90^{0} $$
Done.
answered Mar 16 '16 at 20:33
NarasimhamNarasimham
21.1k62158
$endgroup$
add a comment |
$begingroup$
Hint:
1)$ABC$ and $ABD$ have a right angle and a common angle in $B$.
2)$ABC$ and $ACD$ have a right angle and a common angle in $C$.
answered Mar 16 '16 at 20:22
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
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
$begingroup$
Whether you add angles DAC to BAD or ACD, the result ( sum total) is same, equal to 90 degrees. So they are equal.
$$ angle BAD = angle ACD $$
similarly
$$ angle DBA= angle DAC $$
You said
$$ angle ADB = angle CDA = 90^{0} $$
Done.
answered Mar 16 '16 at 20:33
NarasimhamNarasimham
21.1k62158
$endgroup$
add a comment |
$begingroup$
Whether you add angles DAC to BAD or ACD, the result ( sum total) is same, equal to 90 degrees. So they are equal.
$$ angle BAD = angle ACD $$
similarly
$$ angle DBA= angle DAC $$
You said
$$ angle ADB = angle CDA = 90^{0} $$
Done.
answered Mar 16 '16 at 20:33
NarasimhamNarasimham
21.1k62158
$endgroup$
add a comment |
$begingroup$
Whether you add angles DAC to BAD or ACD, the result ( sum total) is same, equal to 90 degrees. So they are equal.
$$ angle BAD = angle ACD $$
similarly
$$ angle DBA= angle DAC $$
You said
$$ angle ADB = angle CDA = 90^{0} $$
Done.
answered Mar 16 '16 at 20:33
NarasimhamNarasimham
21.1k62158
$endgroup$
Whether you add angles DAC to BAD or ACD, the result ( sum total) is same, equal to 90 degrees. So they are equal.
$$ angle BAD = angle ACD $$
similarly
$$ angle DBA= angle DAC $$
You said
$$ angle ADB = angle CDA = 90^{0} $$
Done.
answered Mar 16 '16 at 20:33
NarasimhamNarasimham
21.1k62158
edited Jan 28 at 1:17
answered Mar 16 '16 at 20:33
NarasimhamNarasimham
21.1k62158
answered Mar 16 '16 at 20:33
NarasimhamNarasimham
21.1k62158
answered Mar 16 '16 at 20:33
NarasimhamNarasimham
21.1k62158
21.1k62158
add a comment |
add a comment |
$begingroup$
Hint:
1)$ABC$ and $ABD$ have a right angle and a common angle in $B$.
2)$ABC$ and $ACD$ have a right angle and a common angle in $C$.
answered Mar 16 '16 at 20:22
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
add a comment |
$begingroup$
Hint:
1)$ABC$ and $ABD$ have a right angle and a common angle in $B$.
2)$ABC$ and $ACD$ have a right angle and a common angle in $C$.
answered Mar 16 '16 at 20:22
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
add a comment |
$begingroup$
Hint:
1)$ABC$ and $ABD$ have a right angle and a common angle in $B$.
2)$ABC$ and $ACD$ have a right angle and a common angle in $C$.
answered Mar 16 '16 at 20:22
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
Hint:
1)$ABC$ and $ABD$ have a right angle and a common angle in $B$.
2)$ABC$ and $ACD$ have a right angle and a common angle in $C$.
answered Mar 16 '16 at 20:22
Emilio NovatiEmilio Novati
52.2k43474
answered Mar 16 '16 at 20:22
Emilio NovatiEmilio Novati
52.2k43474
answered Mar 16 '16 at 20:22
Emilio NovatiEmilio Novati
52.2k43474
answered Mar 16 '16 at 20:22
Emilio NovatiEmilio Novati
52.2k43474
52.2k43474
add a comment |
add a comment |
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1
$begingroup$
It is easiest to show $ABC$ and $ABD$ are similar. They share an angle at $B$.
$endgroup$
– André Nicolas
Mar 16 '16 at 20:21