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Two triangles are similar if a Euclidean similarity maps one onto the other. Their corresponding angles agree, and corresponding side lengths have a common positive ratio.

Specify the correspondence first: $A$ matches $A^{\prime }$, $B$ matches $B^{\prime }$ and $C$ matches $C^{\prime }$. For similar triangles, the common scale factor $k>0$ satisfiesThe ratios compare matching sides in the same order. A Euclidean similarity can also move, turn or reflect the triangle; its orientation on the page is irrelevant. In the Euclidean plane, matching just two interior angles is enough: their sums force the third to match, and equal corresponding angles force the side ratios to agree. This is the angle-angle (AA) criterion for similar triangles. Matching just one angle is insufficient.

Scale the right triangle with vertices $(0,0)$, $(4,0)$ and $(0,3)$ by $k=2$. Its new vertices are $(0,0)$, $(8,0)$ and $(0,6)$, and its side lengths are $6,8,10$. The matching side ratios are $6/3=8/4=10/5=2$. Both triangles have a right angle and the same two other interior angles, approximately $36.87^{\circ }$ and $53.13^{\circ }$. However, their areas are $3\cdot 4/2=6$ and $6\cdot 8/2=24$: doubling lengths multiplies area by four.

Generally, a Euclidean similarity multiplies both a base and its perpendicular height by $k$, so the area becomes $\frac{1}{2}(kb)(kh)=k^2\frac{1}{2}bh$. This explains why an enlarged drawing needs four times as much surface when each length doubles. It also enables indirect measurement: under assumptions of level ground, vertical objects and parallel light rays, an object of height $3$ casting a shadow of length $4$ provides a right triangle. Another object's shadow of length $8$ gives a similar triangle and height $3(8/4)=6$. The shared angles justify the inference; visual resemblance alone does not.

Euclid's Elements VI, Definition 1 describes similarity using corresponding angles and proportional sides. VI.4 establishes the implication from matching angles to matching side ratios, while VI.19 states the squared relation between side ratios and areas. These are documented ancient formulations of the principles used above, without establishing a first discoverer.^{[ref][ref][ref]}

Trigonometric functions turn an angle into a coordinate or a ratio of lengths. Start at $(1,0)$ on the unit circle and rotate counterclockwise through $\theta$ radians: the resulting point isThe cosine records horizontal position and the sine records vertical position. This definition continues around the whole circle, including negative angles, so the signs encode direction rather than requiring a separate rule for each right triangle.

The unit circle picture explains why these functions are useful beyond measuring triangles. The two coordinates of a point rotating at constant angular speed are periodic functions. Scaling the radius to $A$ and using $\theta =\omega t+\varphi$ gives $x=A\cos (\omega t+\varphi )$ and $y=A\sin (\omega t+\varphi )$. A circular motion becomes two simple repeating traces, connecting Euclidean geometry, simple harmonic motion, and Fourier series.

For an acute angle in a right triangle, rescaling its hypotenuse to one recovers the familiar ratios: sine is opposite side divided by hypotenuse, and cosine is adjacent side divided by hypotenuse. The tangent function is their ratio, $\tan \theta =\sin \theta /\cos \theta$, wherever the denominator is nonzero; the secant function, cosecant, and cotangent give the other reciprocal ratios.

The same names extend to complex numbers through the complex exponential:These definitions agree with the circle interpretation for real angles. For complex arguments, the values need not lie between $-1$ and $1$; the literal circle-coordinate picture is restricted to real arguments. ^[ref]

Total-protein Western blot normalization compares a target-band signal with a signal representing many transferred proteins in that lane. It is distinct from measuring one reference protein, and from measuring the target itself. Its denominator is a loading proxy: normalization estimates target abundance relative to detected total protein, rather than target amount per biological cell. Different sample compositions, selective Western blot transfer losses or unequal stain responses can still matter.

For an original calculation, suppose background-corrected target signals $I_A=12$, $I_B=24$ and total-protein signals $P_A=30$, $P_B=60$ are in their respective proportional response ranges. ThenThe invented signals double together, so this calculation supplies no evidence for a higher target fraction. With $I_B=24$ but $P_B=30$, the same calculation gives $2$. These examples assume a common target response and a common total-protein response between lanes, negligible remaining background, and comparable target recovery; they do not equate total-protein staining with target abundance. Multiplying both signals in a lane by the same loading factor cancels in their ratio; a selective target loss does not cancel.

Western blot response saturation means that increasing target amount no longer increases the recorded signal proportionally. It can arise before imaging, through limited binding or reagent availability, as well as through detector limits. Shortening exposure can address a detector limit without repairing an already unresponsive assay. Test the target and normalization readouts with a dilution series under the actual assay conditions; merely seeing a gray band does not establish a proportional response.

In a 2013 methods experiment, a biological cell extract containing added purified yeast protein was separated on gels, transferred with a Trans-Blot Turbo to polymer membranes, and read with an enzyme-linked secondary antibody. A ChemiDoc MP imager and film recorded the same dilution-series blots. The target and an abundant reference protein had different usable response ranges; transferred total-protein imaging and the reference bands did not track loading equally well. This demonstrates the need for assay-specific response checks, rather than a universal conversion from darkness to amount. ^[ref]

An illustrative response model separates background $b$, target amount $T$, retained fraction $q$, response scale $S$ and characteristic amount $K$:This is an assumed model, not a fitted Western blot calibration. When $qT\ll K$, $I-b\approx (Sq/K)T$. When $qT\gg K$, $I-b$ approaches $S$. For two otherwise identical lanes, doubling $T$ gives the background-corrected ratioAt $qT=K$ the ratio is $4/3$, despite a true twofold amount change. Thus Western blot normalization cannot recover a lost amount difference merely by dividing two saturated readouts.