Similar triangles and the proof habit
Tonight's idea
Two triangles are similar when one is a scaled copy of the other: same angles, sides in the same ratio \(k\). Any one of these is enough to prove it:
- AA: two angles match (the third then matches automatically).
- SAS: two sides are in the same ratio and the angle between them matches.
- SSS: all three sides are in the same ratio.
Once triangles are similar, every length is in ratio \(k\) and every area is in ratio \(k^2\). A proof is a chain of statement, reason. Here is the model.
Claim. Triangles \(ACD\), \(CBD\) and \(ABC\) are all similar.
Proof. \(\triangle ACD\) and \(\triangle ABC\) share angle \(A\) and each has a right angle: similar by AA. Likewise \(\triangle CBD\) and \(\triangle ABC\) share angle \(B\): similar by AA. Two triangles similar to a third are similar to each other. \(\square\)
Pythagoras in three lines. From \(\triangle ACD \sim \triangle ABC\): \(\dfrac{AC}{AB} = \dfrac{AD}{AC}\), so \(b^2 = AD \cdot c\). From \(\triangle CBD \sim \triangle ABC\): \(a^2 = BD \cdot c\). Add: \(a^2 + b^2 = c\,(AD + BD) = c^2\). \(\square\)
Watch one: a shadow
A 1.8 m person casts a 2.4 m shadow. At the same moment a tree casts a 14.4 m shadow. The sun's rays are parallel, so the angle at the tip of each shadow is the same, and both stand at right angles to the ground. AA: the triangles are similar. So \(\dfrac{\text{tree}}{14.4} = \dfrac{1.8}{2.4}\), and the tree is \(1.8 \times 14.4 / 2.4 = 10.8\) m tall.
Your turn
- Are triangles with sides 6, 8, 10 and 9, 12, 15 similar? Which test, and what is \(k\)? What about 5, 7, 9 and 10, 14, 17?
- A right triangle has \(AC = 6\) and \(BC = 8\), right angle at \(C\), as in the figure. Draw the altitude \(CD\) to the hypotenuse. Using the similar triangles from Tonight's idea, find the two pieces \(AD\) and \(BD\) of the hypotenuse and the altitude \(h\). Check that \(h^2 = AD \cdot BD\).
- Two similar triangles have areas 18 cm² and 50 cm². A side of the smaller one is 6 cm. How long is the corresponding side of the larger?
- Write a proof. \(M\) and \(N\) are the midpoints of \(AB\) and \(AC\) in \(\triangle ABC\). Prove that \(MN \parallel BC\) and \(MN = \tfrac12 BC\). (Hint: show \(\triangle AMN \sim \triangle ABC\) by SAS. Then \(\angle AMN = \angle ABC\), and equal corresponding angles force \(MN \parallel BC\).)
Stretch: a square root with a ruler
Using \(\triangle ACD \sim \triangle CBD\) in the figure, prove \(h^2 = AD \cdot BD\). Then: on a line, mark segments of length 1 and 7 end to end, draw a semicircle on the whole length 8, and raise a perpendicular at the join up to the semicircle. How long is it, and why? (The angle in a semicircle is a right angle.)
Answers
- Yes, by SSS: each side of the second is \(\tfrac32\) of the first, so \(k = \tfrac32\). The second pair: \(10/5 = 2\) and \(14/7 = 2\) but \(17/9 \ne 2\), so not similar.
- Hypotenuse \(c = 10\). From \(b^2 = AD \cdot c\) with \(b = 6\): \(AD = 36/10 = 3.6\). From \(a^2 = BD \cdot c\) with \(a = 8\): \(BD = 6.4\). Sum 10. Area two ways: \(\tfrac12 \cdot 6 \cdot 8 = \tfrac12 \cdot 10 \cdot h\), so \(h = 4.8\). Check: \(3.6 \times 6.4 = 23.04 = 4.8^2\).
- Areas scale by \(k^2\), so \(k^2 = 50/18 = 25/9\) and \(k = 5/3\). The side is \(6 \times \tfrac53 = 10\) cm.
- \(\dfrac{AM}{AB} = \dfrac{AN}{AC} = \tfrac12\) and angle \(A\) is shared, so \(\triangle AMN \sim \triangle ABC\) by SAS with \(k = \tfrac12\). Hence \(MN = \tfrac12 BC\), and \(\angle AMN = \angle ABC\); equal corresponding angles mean \(MN \parallel BC\).
Stretch. In \(\triangle ACD \sim \triangle CBD\), the side opposite angle \(A\) in the first (\(CD = h\)) corresponds to the side opposite angle \(BCD\) in the second (\(BD\)), and \(AD\) corresponds to \(CD\). So \(\dfrac{h}{BD} = \dfrac{AD}{h}\), giving \(h^2 = AD \cdot BD\). The point on the semicircle above the join makes a right angle with the two ends, so the perpendicular is the altitude of a right triangle with hypotenuse pieces 1 and 7: its length is \(\sqrt{1 \times 7} = \sqrt7\).