Perpendicular lines (also known as orthogonal lines) are a pleasure to work with, because they make things 'right'. There are several cases in which having a right angle greatly simplifies the problem. A simple application of right angles is the Pythagorean Theorem, which states that in a right triangle $ABC$, where $c$ is the length of the hypotenuse and $a$ and $b$ are the lengths of the legs, then $a^2+b^2= c^2$. This is a formula that everyone should be familiar with.

Another useful fact about perpendicular lines is that if we know that two triangles are right-angled, then the 'SSA similar triangles' hold. The reason for this, is that the ambiguous Sine Rule case is now ruled out.

In coordinate geometry, another useful property is that 2 lines are perpendicular if and only if the slope of one is the negative reciprocal of the other (ignoring case where slope = 0). Note that graphing would not help with determining perpendicular lines, unless an accurate diagram is drawn to scale.

The idea of orthogonality also extends to other areas of math, especially in Linear Algebra and Vector Spaces. For example, we say that 2 vectors are perpendicular if their dot product is 0, i.e. $\langle x, y \rangle = 0$, which generalizes to $n$ dimensions.

## If $AB$ is perpendicular to $BC$, and $AB$ is perpendicular to $BD$, then points $B, C$ and $D$ lie on the same line.

If points $C$ and $D$ lie on the same side of $AB$, then since $\angle ABC = \angle ABD$, hence $B, C$ and $D$ lie on the same line. This doesn't require the angle to be right. If points $C$ and $D$ lie on opposite sides of $AB$, then since $\angle ABC + \angle ABD = 90^\circ + 90^\circ$, hence $CBD$ is a straight line. $_\square$

## $PQR$ is a right angled triangle with $\angle QPR = 90^\circ$. From $P$, drop a perpendicular to $QR$ intersecting at $S$. Show that triangles $PQR, SQP, SPR$ are similar. Hence show that $SQ \cdot SR = SP^2$.

By angle chasing, $\angle QPS = 90^\circ - \angle PQS = 90^\circ - \angle PQR = \angle QRP$. Similarly, $\angle RPS = 90^\circ - \angle SRP = 90^\circ - \angle QRP = \angle PQR$. Thus, by PPP, these three triangles are similar. By similarity, $\frac {QS}{SP} = \frac {SP}{SR}$, which gives $QS \cdot SR = SP^2$. $_\square$

## Show that $AB \perp CD$ if and only if $AD^2 + BC^2 = AC^2 + BD^2$. There is no requirement that line segments $AB$ and $CD$ intersect.

This problem is better presented using vector notation. Let the origin be $O$ (doesn't matter where). Let the points $A, B, C, D$ have position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$. Then, $AD^2 = \langle a-d, a-d \rangle = |a|^2 + |d|^2 - 2 \langle a,d \rangle$. $BC^2 = \langle b-c, b-c \rangle = |b|^2 + |c|^2 - 2 \langle b,c\rangle$. $AC^2 = \langle a-c, a-c \rangle = |a|^2 + |c|^2 - 2 \langle a,c \rangle$. $BD^2 = \langle b-d,b-d \rangle = |b|^2 + |d|^2 - 2 \langle b,d \rangle$.

Hence, $2 \langle b-a, d-c \rangle = AD^2 + BC^2 - AC^2 - BD^2$, which shows that $AB \perp CD$ if and only if $AD^2 + BC^2 = AC^2 + BD^2$. $_\square$

Note:Students who are unfamiliar with vectors should prove this statement using the Cosine Rule (which is actually what we're using). Slight care has to be taken with regards to where $AB$ intersects $CD$.

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