*Right Triangle Altitude Theorem*and use it to derive

*Pythagorean Theorem*.

Right Triangle Altitude Theorem

We know that two similar triangles have three pairs of

*equal angles*and three pairs of

*proportional sides*. If someone asks you what your favourite example of similar triangles is, what would you say?

For me, it has to be the

*Right Triangle Altitude Theorem*. The theorem is constructed as follows.

First, we draw a right triangle $ABC$ ($\angle B$ is the right angle). Next, we draw the altitude $BH$ and divide the triangle $ABC$ into two smaller right triangles $BHA$ and $BHC$. Do you see that these two smaller triangles, $BHA$ and $BHC$, are similar to the original triangle $ABC$?

We can draw the picture on a paper and use scissor to cut out the shapes as follows.

Let us look at the big triangle $ABC$ and the smaller triangle $AHB$. We can see right away that they have a pair of equal angles $\angle B = \angle H = 90^{o}$. In addition to that, they also share a common angle $A$. So these two triangles are similar.

Again, the equal ratio between the sides of the two triangles $ABC$ and $BHC$: $$\frac{AB}{BH} = \frac{{\bf BC}}{{\bf HC}} = \frac{{\bf AC}}{{\bf BC}}$$ gives us $$CB^2 = CH \times CA$$

Finally, the ratio for the two triangles $AHB$ and $BHC$ $$\frac{{\bf AH}}{{\bf BH}} = \frac{{\bf HB}}{{\bf HC}} = \frac{AB}{BC}$$ gives us the third identity $$HB^2 = HA \times HC$$

The

*Right Triangle Altitude Theorem*consists of these three identities. Let us call them "t

*he left identity*", "

*the right identity*" and "

*the middle identity*".

Right Triangle Altitude Theorem:

left identity: $AB^2 = AH \times AC$right identity: $CB^2 = CH \times CA$and middle identity: $HB^2 = HA \times HC$ |

Pythagorean Theorem

We will now use the

*Right Triangle Altitude Theorem*to give a proof of the*Pythagorean Theorem*.Pythagorean Theorem:In a right triangle $ABC$ with the right angle $B$ we have $$AB^2 + BC^2 = AC^2$$

*Pythagorean Theorem*says that the two squares $ABXY$ and $BCPQ$ have a total area equal to the big square $CAIJ$.

- Using the
*left identity*$AB^2 = AH \times AC = AH \times AI$, we can see that the*square*$ABXY$ has the same area as the*rectangle*$AHMI$.

- The
*right identity*$CB^2 = CH \times CA = CH \times CJ$ shows that the*square*$BCPQ$ has the same area as the*rectangle*$CHMJ$.

So indeed, the

*big square*$CAIJ$ has area equal to the sum of

*two smaller squares*$ABXY$ and $BCPQ$, and we have obtained the

*Pythagorean Theorem*.

Let us stop here for now. In the next post, we will explore Gauss' construction of a regular 17-polygon. We will see the reason why the

*Right Triangle Altitude Theorem*makes it possible for this construction. Hope to see you again then.

*Homework.*

1. Prove that if two triangles have two pairs of equal angles then all their three pairs of angles are equal.

2. Write about your favourite example of similar triangles.

3. Use trigonometry to prove the

*Right Triangle Altitude Theorem*.

4. Given three line segments of length $r$, $r a$ and $r b$. Use compass and straightedge to construct

- a line segment of length $r (a+b)$
- a line segment of length $r (a-b)$
- a line segment of length $r (ab)$
- a line segment of length $r (a/b)$
- a line segment of length $r \sqrt{ab}$

5. Given a line segment of length $r$. Using compass and straightedge, what kind of line segments can we construct?

6. Go to google.com and search about compass and straightedge construction.

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