### Video Transcript

In this video, weβll learn how to
use the right triangle altitude theorem, also known as the Euclidean theorem, to
find a missing length. This theorem is a useful tool to
rewrite expressions involving the lengths of sides in a right triangle with a
projection from the right angle onto the hypotenuse. In particular, it allows us to
determine the lengths of sides in a right triangle given two of the lengths. Our plan of action in this video is
we begin with a right triangle, adding squares to each side of the triangle, and use
projections and the properties of congruent triangles to derive the Euclidean
theorem. Weβll then use this theorem,
together with the Pythagorean theorem, to derive a corollary. Weβll then see how the theorem and
the corollary can be applied in some examples.

To derive the Euclidean theorem, we
begin with a right triangle π΄π΅πΆ with right angle at π΄. We then project π΄ onto the side
πΆπ΅ and call this point π· as shown. The line segment π΄π· is
perpendicular to side πΆπ΅. We now add to the diagram three
squares given by each side of triangle π΄π΅πΆ and label the vertices of these
squares as shown. We next want to continue our
projection from π΄ to the side πΆπ΅ down to the side πΊπ». And we then add the line πΆπΉ and
π΄πΊ to our diagram as shown. And we want to show now that
triangle πΆπ΅πΉ and triangle πΊπ΅π΄ are congruent. We can do this by first noting that
the measure of angle πΆπ΅πΉ is equal to the measure of angle π΄π΅πΊ. And this is true because these are
both right angles added to angle π΄π΅πΆ.

Now we know that π΅πΆ is equal to
π΅πΊ since theyβre the sides of a square and similarly that π΄π΅ is equal to
π΅πΉ. And since our triangles have two
sides of the same length, including the same angle measure, this is enough to show
that the triangles are congruent.

Next, we know that the area of a
triangle is half the length of the triangleβs base multiplied by the triangleβs
perpendicular height. Applying this to our triangle
πΊπ΅π΄, which is an obtuse triangle, and choosing π΅πΊ as the base, we note that
π·π΅ is actually our perpendicular height. So the area of triangle π΄π΅πΊ is
one over two multiplied by π΅πΊ times π΅π·. Thatβs one over two times the base
times the height. Now we know that π΅πΊ times π΅π· is
the area of the rectangle π΅π·πΎπΊ. Hence, the area of our triangle
π΄π΅πΊ is one over two times the area of the rectangle π΅π·πΎπΊ.

Similarly, we can see choosing the
base of triangle π΅πΆπΉ to be π΅πΉ that the area of our triangle π΅πΆπΉ is one over
two multiplied by π΅πΉ times π΄π΅. Thatβs where π΄π΅ is the
perpendicular height of the triangle. And we know that in this case π΅πΉ
multiplied by π΄π΅ is the area of the square π΄πΈπΉπ΅ so that the area of our
triangle π΅πΆπΉ is one over two times the area of the square π΄πΈπΉπ΅. Now, recalling that our two
triangles π΄π΅πΊ and π΅πΆπΉ are congruent, they must have the same area. And this means that the square
π΄πΈπΉπ΅ must have the same area as the rectangle π΅π·πΎπΊ.

So now, if we equate the
expressions for their areas, thatβs π΅π΄ squared is equal to π΅π· multiplied by
π·πΎ. π΅π΄ squared is the area of our
square, and π΅π· multiplied by π·πΎ is the area of the rectangle, noting further
that the side π·πΎ has the same length as the side πΆπ» and that πΆπ» has the same
length as πΆπ΅. Since theyβre the sides of a
square, we have that π΅π΄ squared is equal to π΅π· multiplied by π΅πΆ. Making a note of this, in the same
way, we can show that πΆπ΄ squared is equal to πΆπ· multiplied by πΆπ΅. So now making some space, we can
summarize our results in the Euclidean theorem. This says if triangle π΄π΅πΆ is a
right triangle at π΄ with a projection to π· as shown, then π΅π΄ squared is equal to
π΅π· multiplied by π΅πΆ and πΆπ΄ squared is equal to πΆπ· multiplied by πΆπ΅.

We can also use the Euclidean
theorem, together with the Pythagorean theorem, to show another useful result. Now recalling that the Pythagorean
theorem tells us that the hypotenuse of a right angle triangle squared is equal to
the sum of the squares of the other two sides. For our triangle π΄π΅π·, this gives
us π΄π· squared plus π΅π· squared is π΄π΅ squared. And subtracting π΅π· squared from
both sides, we have π΄π· squared is π΄π΅ squared minus π΅π· squared. Now, using our Euclidean theorem,
we substitute π΄π΅ squared, that is, π΅π΄ squared, which is π΅π· multiplied by
π΅πΆ. So we have π΄π· squared is equal to
π΅π· multiplied by π΅πΆ minus π΅π· squared. And taking out a common factor of
π΅π·, on our right-hand side, we have π΅π· multiplied by π΅πΆ minus π΅π·.

And noting that the side length
π΅πΆ is the same as π΅π· plus πΆπ·, subtracting π΅π· from both sides gives us πΆπ·
is equal to π΅πΆ minus π΅π·. And since π΅πΆ minus π΅π· is the
second term on our right-hand side, we have π΄π· squared is π΅π· multiplied by
πΆπ·. And so our corollary to the
Euclidean theorem is that if π΄π΅πΆ is a right triangle at π΄ with projection to π·
as shown, then π΄π· squared is π΅π· multiplied by πΆπ·.

Itβs worth noting that thereβs an
alternative and equivalent way of looking at these two results. This is the Euclidean leg rule,
where again we have a right triangle π΄π΅πΆ with right angle at π΄. And this is projected onto the
hypotenuse πΆπ΅ to the point π·. And this tells us if the leg and
the part are as shown, then the hypotenuse divided by the leg is equal to the leg
divided by the part. That is, π΅πΆ over π΅π΄ is equal to
π΅π΄ over π΅π·. And if we multiply both sides by
π΅π΄ and π΅π·, this gives us the first part of our Euclidean theorem: π΅πΆ
multiplied by π΅π· is π΅π΄ squared.

Now, if we change our leg and part
to the triangle π΄π·πΆ as opposed to π΄π·π΅, again we have hypotenuse over leg is
equal to leg over part. That is, π΅πΆ over πΆπ΄ is equal to
πΆπ΄ over πΆπ·. And we have the second part of our
Euclidean theorem: π΅πΆ multiplied by πΆπ· is πΆπ΄ squared. And similarly, for our second
result, the corollary, also called the altitude rule, if the altitude is the
perpendicular height π΄π· and the hypotenuse is split into left and right parts,
thatβs πΆπ· and π·π΅, we have left over altitude is equal to altitude over
right. And this gives us our corollary
πΆπ· multiplied by π΅π· is π΄π· squared. And what this tells us is that the
altitude or height of the right triangle squared is the product of the lengths of
the segments of the hypotenuse when itβs split by the altitude.

So now letβs look at an example of
applying the Euclidean theorem to find a missing side length in a right
triangle.

Find the length of the line segment
π΄π·.

To find the length π΄π·, we first
note that π· is the projection of π΅ onto π΄πΆ and that the triangle π΄π΅πΆ is a
right triangle at π΅. And now recalling the Euclidean
theorem, this tells us that π΄π΅ squared is π΄π· multiplied by π΄πΆ. That is, the side length π΄π΅
squared is the product of the segment π΄π· with the length of the hypotenuse
π΄πΆ. Weβre given that π΄π΅ is 34
centimeters and that the hypotenuse π΄πΆ is 40 centimeters. And substituting these values into
the Euclidean theorem, this gives us 34 squared is π΄π· multiplied by 40. Dividing through by 40 and
rearranging, this gives us π΄π· is 34 squared over 40, that is, 1156 divided by 40,
which is 28.9. The line segment π΄π· is therefore
28.9 centimeters.

Letβs now see another example of
applying the Euclidean theorem to determine the length of a side in a right
triangle.

Determine the length of line
segment π΄π·.

We note that π· is the projection
of π΄ onto the line π΅πΆ and that the triangle π΄π΅πΆ is a right triangle at π΄. And we recall that the corollary to
the Euclidean theorem, that is, the altitude rule, tells us that π΄π· squared is
equal to π΅π· multiplied by πΆπ·. This tells us that the altitude or
height of the right triangle squared is the product of the lengths of the segments
of the hypotenuse when it is split by the altitude.

Weβre given that the two segments
are 2.5 centimeters, thatβs πΆπ·, and 6.4 centimeters, thatβs π΅π·. And substituting these values in,
we have π΄π· squared is 6.4 multiplied by 2.5. This evaluates to 16, so π΄π·
squared is 16. Now, taking the square root on both
sides of the equation, noting that π΄π· is a length and so is nonnegative, we get
π΄π· is the square root of 16, which is four. The length of π΄π· is therefore
four centimeters.

In our next example, weβll apply
both parts of the Euclidean theorem to determine a missing side length in a right
triangle.

In the figure shown, if ππΏ is
equal to 40 and ππΏ is 30, what is the length of ππ?

We note first that πΏ is the
projection of π onto ππ. And the triangle πππ is a right
triangle at π. We can then recall that the
corollary to the Euclidean theorem tells us that ππΏ squared is πΏπ multiplied by
ππΏ. Now, since we know the lengths of
ππΏ and ππΏ, that is, 40 and 30 centimeters, respectively, this will allow us to
find πΏπ. We can then apply the Euclidean
theorem to find ππ. So now, substituting in the given
lengths to the corollary of the Euclidean theorem, which is also known as the
altitude rule, we have 30 squared is πΏπ multiplied by 40. Dividing through by 40 and
rearranging, this gives us πΏπ is equal to 30 squared divided by 40, that is, 900
divided by 40, which is 22.5.

So making a note of this and
clearing some space, we can now use the Euclidean theorem to note that ππ squared
is equal to πΏπ multiplied by ππ. Remember, πΏπ is the shortest
segment of the hypotenuse, which is 22.5 centimeters, and ππ is the length of the
hypotenuse of the triangle πππ. We know that the length of the
hypotenuse ππ is the sum of the two segments, that is, 40 plus 22.5. And thatβs equal to 62.5. Now, substituting these values into
our Euclidean theorem, thatβs ππ squared is equal to πΏπ multiplied by ππ, we
have πΏπ is 22.5 multiplied by ππ, which is 62.5. And this evaluates to 1406.25. And now taking the positive square
root on both sides, since ππ is a length and so must always be positive, we have
that ππ is 37.5 centimeters.

Itβs worth noting that we could
also have found this length using the Pythagorean theorem with the right triangle
ππΏπ. This gives us ππ squared is equal
to πΏπ squared plus ππΏ squared. That is, ππ squared is 22.5
squared plus 30 squared, which is 1406.25 as before. And so again, taking the positive
square root, we have ππ is 37.5. And so the length ππ is 37.5
centimeters.

In our final example, weβll apply
the Euclidean theorem and the Pythagorean theorem to determine the area of the right
triangle from a given diagram.

Calculate the area of triangle
π΄π΅π·.

We begin by recalling that the area
of a triangle is half the length of the triangleβs base multiplied by the triangleβs
perpendicular height. And since π΄π΅π· is a right
triangle at π·, we can choose π΅π· as the base and then π΄π· as the perpendicular
height. The area of the triangle is then
one over two times π΅π· multiplied by π΄π·. We can find the length π΄π· using
the Euclidean theorem and then the length π΅π· using the Pythagorean theorem.

If we first note that π· is the
projection of π΅ onto the side π΄πΆ and that the triangle π΄π΅πΆ is a right triangle
at π΅, then the Euclidean theorem tells us that π΄π΅ squared is equal to π΄πΆ
multiplied by π΄π·. Substituting now π΄π΅ is 44
centimeters and π΄πΆ is 55 centimeters, we have 44 squared is 55 multiplied by
π΄π·. And dividing through by 55, we have
44 squared over 55 is π΄π·, which evaluates to 35.2 centimeters. And we now have the lengths of two
sides of our right triangle π΄π΅π·. And so we can use the Pythagorean
theorem to find a third length, that is, the length π΅π·.

Applied to our triangle, the
Pythagorean theorem tells us that π΄π΅ squared is π΄π· squared plus π΅π·
squared. And substituting π΄π΅ is 44
centimeters and π΄π· is 35.2 centimeters, we have 44 squared is 35.2 squared plus
π΅π· squared. And subtracting 35.2 squared from
both sides and rearranging, this gives us π΅π· squared is 44 squared minus 35.2
squared. Our right-hand side evaluates to
696.96. And taking the positive square root
on both sides, since π΅π· is a length, thatβs the square root of 696.96, which is
equal to 26.4. So π΅π· is 26.4 centimeters.

So now, we can substitute π΄π· is
35.2 centimeters and π΅π· is 26.4 centimeters into our formula for the area. This gives us the area of triangle
π΄π΅π· is one over two times 26.4 multiplied by 35.2. This evaluates to 464.64, and hence
the area of triangle π΄π΅π· is 464.64 square centimeters.

Letβs now complete this video by
recapping some of the key points weβve covered. The Euclidean theorem tells us that
if π΄π΅πΆ is a right triangle at π΄ with projection to π· on the line πΆπ΅, then
π΅π΄ squared is π΅π· multiplied by π΅πΆ and πΆπ΄ squared is πΆπ· multiplied by
πΆπ΅. There is a useful corollary to the
Euclidean theorem that gives us π΄π· squared is equal to π΅π· multiplied by
πΆπ·. And this is also called the
altitude rule. Sometimes we need to apply both the
Euclidean theorem and its corollary. We may also need to apply these in
conjunction with the Pythagorean theorem to solve problems.