Similar triangles

Theorem. If two triangles are equiangular, their corresponding sides are in proportion, so the triangles are similar. (Reason: equiangular $△$s, or $|||$ $△$s once the similarity is proved)

Proof (examinable). Given $△ABC$ and $△DEF$ with $\hat{A}=\hat{D}$, $\hat{B}=\hat{E}$, $\hat{C}=\hat{F}$. Mark $G$ on $AB$ and $H$ on $AC$ with $AG=DE$ and $AH=DF$. Then $△AGH\equiv △DEF$ (SAS), so $\widehat{AGH}=\hat{E}=\hat{B}$. These are corresponding angles, so $GH\parallel BC$. By the proportionality theorem, $\frac{AB}{AG}=\frac{AC}{AH}$, which gives $\frac{AB}{DE}=\frac{AC}{DF}$. The same method gives the third ratio.

To prove two triangles similar, it is enough to show two pairs of equal angles. The third pair follows from the angle sum.

If $△ABC|||△DEF$, then

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

Order matters. Write the vertices so that equal angles match: $A$ with $D$, $B$ with $E$, $C$ with $F$. Then the ratios come straight from the names.

Converse. If the sides of two triangles are in proportion, the triangles are similar. (Reason: sides of $△$ in prop)

Example. $△ABC|||△DEF$ with $AB=6$, $BC=9$ and $DE=4$. Then $\frac{6}{4}=\frac{9}{EF}$, so $EF=6$.

In right-angled $△ABC$ with $\hat{B}=90°$, draw the altitude $BD\perp AC$. Then

$△ABD|||△ACB|||△BCD$

This gives three useful results:

• $A{B}^2=AD\cdot AC$
• $B{C}^2=CD\cdot CA$
• $B{D}^2=AD\cdot DC$

Adding the first two: $A{B}^2+B{C}^2=AC(AD+DC)=A{C}^2$, which proves Pythagoras.

Proof that the triangles are similar (examinable). In $△ABD$ and $△ACB$: $\hat{A}$ is common and $\widehat{ADB}=\widehat{ABC}=90°$, so $△ABD|||△ACB$ (equiangular $△$s). In $△BCD$ and $△ACB$: $\hat{C}$ is common and $\widehat{BDC}=\widehat{ABC}=90°$, so $△BCD|||△ACB$ (equiangular $△$s). Then $△ABD|||△BCD$ since both are similar to $△ACB$. The IEB lists this as an examinable proof: the perpendicular from the right angle to the hypotenuse divides the triangle into two triangles similar to each other and to the original triangle.

Example. If $AD=4$ and $DC=9$, then $B{D}^2=36$, so $BD=6$, and $A{B}^2=4\times 13=52$, so $AB=2\sqrt{13}$.

In circle riders, the equal angles you need for similarity usually come from circle theorems:

• angles in the same segment
• the tan-chord theorem
• exterior angles of cyclic quadrilaterals
• vertically opposite angles where chords cross

Example: intersecting chords. Chords $AB$ and $CD$ cross at $P$. In $△APC$ and $△DPB$:

• $\hat{A}=\hat{D}$ ($\angle$s in the same seg, subtended by $CB$)
• $\widehat{APC}=\widehat{DPB}$ (vert opp $\angle$s)

So $△APC|||△DPB$ (equiangular $△$s), and $\frac{AP}{DP}=\frac{PC}{PB}$, giving $AP\cdot PB=CP\cdot PD$.

A typical question asks you to prove two triangles similar and then "hence" prove a product of lengths. The product almost always comes from cross-multiplying the ratio.