Theorems about Similar Triangles
1. The Side-Splitter Theorem
If ADE is any triangle and BC is drawn parallel to DE, then AB/BD = AC/CE
To show this is true, draw the line BF parallel to AE to complete a parallelogram BCEF:
Triangles ABC and BDF have exactly the same angles and so are similar (Why? See the section called AA on the page How To Find if Triangles are Similar.)
- Side AB corresponds to side BD and side AC corresponds to side BF.
- So AB/BD = AC/BF
- But BF = CE
- So AB/BD = AC/CE
The Angle Bisector Theorem
If ABC is any triangle and AD bisects the angle BAC, then AB/BD = AC/DC
To show this is true, we can label the triangle like this:
- Angle BAD = Angle DAC = x°
- Angle ADB = y°
- Angle ADC = (180 - y)°
By the Law of Sines in triangle ABD:
sin x°/BD = sin y°/AB
So AB × sin x° = BD × sin y°
And so:
AB/BD = sin y°/sin x°
By the Law of Sines in triangle ACD:
sin x°/DC = sin(180 - y)°/AC
So AC × sin x° = DC × sin(180 - y)°
So AC/DC = sin(180 - y)°/sin x°
But sin(180 - y)° = sin y°
And so:
AC/DC = sin y°/sin x°
Combining AC/DC = sin y°/sin x° with AB/BD = sin y°/sin x° gives:
AC/DC = sin y°/sin x° = AB/BD
So AB/BD = AC/DC
In particular, if triangle ABC is isosceles, then triangles ABD and ACD are congruent triangles
And the same result is true:
AB/BD = AC/DC
3. Area and Similarity
If two similar triangles have sides in the ratio x:y,
then their areas are in the ratio x^{2}:y^{2}
Example:
These two triangles are similar with sides in the ratio 2:1 (the sides of one are twice as long as the other):
What can we say about their areas?
The answer is simple if we just draw in three more lines:
We can see that the small triangle fits into the big triangle four times.
So when the lengths are twice as long, the area is four times as big
So the ratio of their areas is 4:1
We can also write 4:1 as 2^{2}:1
The General Case:
Triangles ABC and PQR are similar and have sides in the ratio x:y
We can find the areas using the formula from Finding the area of a triangle that has no right angle.
ABC's Area = ½bc sin A
PQR's Area = ½qr sin P
And we know the lengths of the triangles are in the ratio x:y
q/b = y/x, so: q = by/x
and r/c = y/x, so r = cy/x
Also, since the triangles are similar, angles A and P are the same:
A = P
We can now do some calculations:
Area of triangle PQR: | ½qr sin P | |
Put in "q = by/x", "r = cy/x" and "P=A": | ½(by/x)(cy/x) sin A | |
Combine (by/x) and (cy/x): | ½(bycy/xx) sin A | |
Simplify: | ½(bcy^{2}/x^{2}) sin A | |
Rearrange: | y^{2}/x^{2} × ½(bc) sin A | |
Which is: | y^{2}/x^{2 } × Area of Triangle ABC |
So we end up with this ratio:
Area of triangle ABC : Area of triangle PQR = x^{2 }: y^{2}