Corresponding Angles?

I wanted to prove that α=β\alpha = \beta here: Given ABCD\overline{AB} \parallel \overline{CD}.

The first thing needed was to prove that the distance between the intercepts (rather the vertical distance) remained constant for any xx value. This can be done using systems of equations easily: y1=mx+b1y2=mx+b2y_{1} = mx + b_{1}\\y_2 = mx + b_{2} You subtract mxmx on both sides and are left with 0x0x, which means that for an xx value, you will always have a result of the difference between b1b_{1} and b2b_{2}. This proves that, for any domain, the distances between the two lines will remain the same.

Discounting, for a moment, an extraneous or general transversal, let us start with two cases perpendicular to the x-axis on a Cartesian grid: We have already proven above that ABDC\overline{AB} \cong \overline{DC}, now we can show that BCAD\overline{BC} \cong \overline{AD} with the Pythagorean theorem if we construct a right triangle making the respective line segments the hypotenuses:

x2+y2=BC2=AD2x^{2} + y^{2} = \overline{BC}^{2} = \overline{AD}^2 Taking the square root, you can see that they are also congruent.

Next, we can show that the perpendicular case holds true: To best show this, you can construct a line segment BDABDCDB\overline{BD} \Rightarrow \bigtriangleup{ABD} \cong \bigtriangleup{CDB} With the predetermined congruence and the SSS postulate, this can be show to be true. This means also that: mDABmBCDm \angle{DAB} \cong m \angle{BCD} By the definition of a rectangle, mECB=90θmCBE=θm \angle{ECB} = 90 - \theta \Rightarrow m \angle{CBE} = \theta.

Taking away everything now immaterial, we see that the corresponding angles theorem is true for the perpendicular case:

How do we now establish the theorem for all cases? Recall that we have already shown that for any value, we can construct a congruent parallelogram. Becuase the Cartesian coordinate system is relative, we can center it so that our transversal touches the bottom left and top right corners of the parallelogram for any transversal. Here is an example: We can again construct the parallelogram as before with the special case: Established previously, we know ABDC\overline{AB} \cong \overline{DC} and ADBC\overline{AD} \cong \overline{BC}. Rather perspicaciously, we know also ACAC\overline{AC} \cong \overline{AC}. And again, by SSS, mBACmDCAm \angle{BAC} \cong m \angle DCA. Simple Euclidean geometry will tell you, then, that mDCAmECFm \angle{DCA} \cong m \angle{ECF}. The product is realized when we again clear the excess:

QEDQED

Note by Drex Beckman
3 years, 1 month ago

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You could just draw line BDBD.Then we have a+180b=180,ab=0,a=ba + 180 - b = 180, a - b = 0, a = b

Samuel Sturge - 3 weeks, 2 days ago

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