One-line three equal angles (K-shape similarity)

When to use

• Three equal angles, all on the same side of a single line (most common is $90°$, but any $\alpha$ works)
• The figure shows a "polyline / Z-shape / K-shape": a moving point $P$ on a line, with slanted segments on each side meeting the line at the same angle
• Find segment-length ratios; or pin down the moving point that "makes two segments proportional"
• In rectangle / square problems with "moving-point polylines": the right angle at the bend is the equal angle on the line

Core move

Three equal angles on one line ⇒ the two adjacent triangles are AA-similar: use corresponding ratios $\dfrac{AB}{PC} = \dfrac{BP}{CD}$ to solve for unknown lengths.

Construction

Let line $\ell$ contain $B, P, C$ in order, with $A, D$ on the same side of $\ell$, and $\angle ABP = \angle APD = \angle DCP = \alpha$:

1. Use the straight angle at $P$: the three angles at $P$ sum to $180°$, so $\angle APB + \alpha + \angle DPC = 180°$, giving $\angle APB + \angle DPC = 180° - \alpha$.

   

2. Chase triangle interior angles: in $\triangle ABP$, $\angle BAP + \angle APB + \alpha = 180°$, so $\angle BAP = 180° - \alpha - \angle APB = \angle DPC$.

3. AA similarity: $\angle ABP = \angle PCD = \alpha$ and $\angle BAP = \angle DPC$, so $\triangle ABP \sim \triangle PCD$.

4. Write the ratio: matching corresponding sides — $\dfrac{AB}{PC} = \dfrac{BP}{CD} = \dfrac{AP}{PD}$.

   

5. Substitute and solve: typically three of the four lengths are given and you solve for the fourth, or you set up an equation that locates $P$.

Why it works

"Three equal angles on a line" is equivalent to "the slanted-line directions on the two sides correspond pairwise" — the essential condition for a similarity transformation. The straight angle at $P$ "transports" the third equal angle across, completing the AA criterion. The mechanism is the same as alternate / corresponding angles, just with the angle equal to a generic $\alpha$ rather than $0°$ or $90°$.

Worked examples

• Rectangle $ABCD$, $P$ on $BC$ with $\angle APE = 90°$ ($E$ on $CD$): find the relationship between $BP$ and $CE$ (the $\alpha = 90°$ version)
• Equilateral triangle folded so the apex lands on the base: three $60°$ angles align ⇒ the two side triangles are similar

Variants / generalizations

• Degenerate to right angles ($\alpha = 90°$): commonly called the "K-shape" or "one-line three right angles" — the most familiar middle-school version.
• Bow-tie / X-shape: two lines intersect at a point with four equal angles formed by vertical pairs — similar mechanism, gives "vertical-angle similarity".
• Add the isosceles condition (e.g. $AB = BP$): similarity upgrades to congruence, and gives the geometric meaning of "rotation by $\alpha$ around the bend".