Question Video: The Right-Angled Triangle Altitude Theorem | Nagwa Question Video: The Right-Angled Triangle Altitude Theorem | Nagwa

# Question Video: The Right-Angled Triangle Altitude Theorem Mathematics • Second Year of Preparatory School

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What does (π΄πΆ)Β² equal to?

03:18

### Video Transcript

What does π΄πΆ squared equal to?

Weβre asked what the square of the length of side π΄πΆ in the triangle shown is equal to. And to work this out, weβre going to use areas. In fact, weβll see that this is actually one-half of the Euclidean theorem for right triangles.

We begin by redrawing triangle π΄π΅πΆ, adding a square to each of its sides, and labeling the vertices of these squares as shown. Our next step is to project π΄ onto side πΊπ», meeting it at point πΎ, and then adding the lines π΅πΈ and π΄πΊ to the diagram as shown. We want to now show that triangles π΅πΆπΈ and πΊπΆπ΄ are congruent. And from this, via equating some areas, weβll find an expression for π΄πΆ squared.

The first thing we can note is that angles π΅πΆπΈ and πΊπΆπ΄ are congruent. This is because theyβre both right angles added to angle π΄πΆπ΅. Now, we also know that side πΆπ΅ is equal to side πΆπΊ and that side π΄πΆ is equal to side πΆπΈ. So we see that triangles π΅πΆπΈ and πΊπΆπ΄ are congruent since they have at least two congruent sides with the included angle between those sides also congruent.

Now, we know that the area of any triangle is half the base times the perpendicular height. So, now applying this to triangle πΊπΆπ΄, where we choose πΊπΆ as the base and where πΆπ· is the perpendicular height, we have that the area of triangle πΊπΆπ΄ is equal to one-half of πΊπΆ times πΆπ·. We know also that πΊπΆ times πΆπ· is equal to the area of the rectangle πΆπ·πΎπΊ. So we have the area of triangle πΊπΆπ΄ equals half the area of the rectangle πΆπ·πΎπΊ.

So, now making some space and making a note of this, we can look similarly at triangle π΅πΆπΈ. This time, we choose our base as side πΈπΆ, and the perpendicular height is πΆπ΄. Then, the area of triangle π΅πΆπΈ equals one-half πΈπΆ times πΆπ΄. Now we note that πΈπΆ times πΆπ΄ is the area of the square π΄πΆπΈπΉ. And we can observe also that since triangles πΊπΆπ΄ and π΅πΆπΈ are congruent, they must have the same area. This, in turn, means that the rectangle πΆπ·πΎπΊ must have the same area as the square π΄πΆπΈπΉ.

Now, the area of rectangle πΆπ·πΎπΊ equals πΆπ· times π·πΎ. And π·πΎ equals πΆπΊ, which is equal to πΆπ΅. So the area of rectangle πΆπ·πΎπΊ is equal to πΆπ· times πΆπ΅. Similarly, the area of the square π΄πΆπΈπΉ equals π΄πΆ times πΈπΉ. And since πΈπΉ equals π΄πΆ, this squareβs area is equal to π΄πΆ squared. So we have the area of the square π΄πΆπΈπΉ equal to the area of the rectangle πΆπ·πΎπΊ. And so, π΄πΆ squared is equal to πΆπ· multiplied by πΆπ΅, which is actually one part of the Euclidean theorem for right triangles.

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