similar triangles, same me....

... from this obsession of diving deep into shallow waters, somebody save me :-)

Usually when similar triangles are discussed, the texts start off by stating their properties or maybe exemplify a bit, and then take off into describing various tricks to deduce similarity between given set of triangles.

Of course, if you keep drawing to scale or use intuition, you would reckon that similar triangles are like aspect ratio locked pictures, you can scale their size up or down but you wouldn't be able to distort them.

But me being me stopped and asked, why is it that if the angles within two triangles are equal, the ratios of their corresponding sides must be equal? pretending to forget both the concept of slope in co-ordinate geometry as well as trigonometry for a moment. Because, obviously circular references are bad ;-)

As per usual practice, looked up the net, but on this one, could not find an explanation satisfactory to my taste and here goes one that i write for myself.

So, lets first set our task up with an illustration:

This arrangement of two triangles ABC and DEF with labeling of the internal angles is our starting point, but not much helpful, so lets superimpose the triangles and check if we see a pattern:

Hmm, not much helpful either but we can see that what we need to find out is: how does the Δa of change on 'a' relate to the corresponding Δb of change in 'b'.

This would get easier if our 'Δa' is actually an integral multiple of 'a'. By shrinking the units, we can extend the relationship to fractional increase as well, we'll get to this at the very end.

So, lets start with the simplest possible extension, that is, lets double the length of the side that measures 'a' :

We have the smaller triangle AED with side lengths AE = a, ED = c, and DA = b.

And we have the bigger triangle ABC with side length AB = 2*a, but we don't really know the lengths of other sides BC and CA yet.

However, we do know that CA is an extension of DA(=b), so if we can find the length of CD, we can sum it up with 'b' to find the length of CA.

That is simple to do as illustrated in the figure below:

On the extended triangle, we draw 2 more lines to make our analysis easier. First, we draw a line EF parallel to AC intersecting point E, and then we draw a line DF parallel to AB going through point D.

With these two lines in place, we observe two parallelograms.

 

The brown parallelogram AEFD reflects the length 'b' from side DA to side EF, and then the orange parallelogram DEFC reflects the length 'b' from side EF onto side CD.

Thus, we can conclude that length of CD = 'b', and hence the length of the side CA in the bigger triangle is indeed CD+DA = b+b = '2*b'.

If you are unsure as why the two extra lines we drew form a parallelogram, here some explanation: consider the triangle at right bottom, EBF. ∠FEB = ∠DAE = α; AE = EB = a; and ∠EBF = ∠AEF = β. 

Using Angle-Side-Angle theorem of triangle congruence, we can conclude, triangle EBF is congruent to triangle AED.

And hence, EF = DA = 'b' and it follows that FD = AE = 'a'.

 

Lets now look at the below figure to understand the length of the other side BC:

Here, the brown parallelogram DEFC reflects the length 'c' from side DE onto side FC.

The purple parallelogram EBFD reflects the length 'c' from the side DE onto side BF.

So, we have FC = c and BF = c

Together the length of the other side of the bigger triangle BC = BF + FC = 2*c

Thus by relying on a simple fact that the opposite sides of a parallelogram are equal, we just showed that when a triangle is enlarged by keeping the internal angles fixed, and doubling the length of one side, the length of other two sides double up too!

Lets now consider the situation where we extend one side by some 'n' times its original size without changing internal angles:

 

We can draw lines parallel to side with length c at each of the n divisions on the side that measures (n+1) times 'a' and get something as below:

Using same logic as we did when the side doubled, we can show that the AB/AE = BC/ED = CA/DA = 'n'

What about the extensions that are not integral multiples? Lets sort that out by means of an example. Consider we have a small triangle AED, that we are enlarging into a bigger triangle ABC and that AE = 5 and AB = 7.9 (instead of some multiple of 5)

We can divide the side AE of the AED triangle into 50 equal units, draw lines parallel to ED passing through each of those 50 points, and there we have 50 triangles in ascending order of their size but with the same internal angles. Lets us denote the sides of the smallest of these triangles with unit length on the AE segment as Δa, Δc, and Δb respectively. And the corresponding sides AE, ED, and DA be 'a', 'c', and 'b' respectively

From our earlier analysis we can show that a/Δa = c/Δc = b/Δb = 50  ----- (1)

Similarly, if we consider the bigger triangle ABC, we can create 79 triangles of unit side length on the side corresponding to AB and if we consider AB, BC, and CA as a`, c`, and b` respectively, we can show that a`/Δa = c`/Δc=b`/Δb = 79  ---- (2)

From (1) and (2) a`/a = b`/b = c`/c = 79/50

And so much goes for the lack of intuition :-)

Midpoint of a Straight line

Which of the below two is easier to convince yourself of?

a) the angle-side-angle theorem of triangle congruence

b) relative proportions of a line remain consistent with its projection (shadow).

i know it ought to be b).... but on a muddled up day...  you invent some muddle up logic ...

Don't ask me why but i had to calculate the midpoint of a straight line between two points : A (x1, y1) and B: (x2, y2) on the cartesian plane :-)

There is the well known formula for the midpoint, if we call it C : ( (x1+x2)/2, (y1+y2)/2 )

Though it is visually convincing to see that the x co-ordinate and y co-ordinate of the mid point C fall midway between the x and y co-ordinates of the point A and B, i was trying to build some logical reasoning behind it, if not the proof. Looked around a bit online, but couldn't really find a satisfactory explanation and hence wrote one for myself.

Let's take a look at the below example with points A (1, 4) and B (7, 2) whose midpoint lies at C (4, 3)

Its easy to see this by taking the projections of points A, B, C on X and Y axes and realizing that the X-cord of C is exactly half way between the X-coords of A and B. And likewise with the Y-Coords...

But it seems difficult to put this into a logically tight expression. 

How else can we deduce the midpoint in a sound way? Lets define the midpoint again.. It is a point that lies exactly half way between the end points of a finite line segment.

So, if we consider the AB as the line segment, and C as the midpoint, length of AC is same as length of CB. Lets denote this length AC = CB = m. 

Let's look at the above same picture but with a different perspective now, repurposed as below:

The triangles AOC, and CQB are congruent because:

a)  ∠AOC = ∠CQB = 90

b) ∠OCA = ∠QBC  Why? Because, A straight line (AB) makes the same angle with two parallel lines (OC and QB).

c) Length of sides AC = CB = m 

d) Additionally, ∠CAO = ∠BCQ (from a and b)

So, using angle-angle-side or angle-side-angle theorems, we can conclude that triangles AOC and CQB are congurent.

Hence:

1) side AO = CQ Which means Y-coord of C is exactly halfway between the Y-cords of points A and B, given that Q's Y-Coord is same as B's Y-Coord.

2) side OC = QB. Which means X-coord of C is exactly halfway between the X-coords of points A and B, given that O and A have same X-coords, and C and Q have same X-coords.

Ofcourse, here we took help of angle-side-angle theorem for our deduction.