Congruent Figures


In mathematics, we say two things are congruent if they are the same, up to some kind of transformation.

For example, 12=24 \frac{1}{2} = \frac{2}{4} because they both represent the value 0.5 0.5, though they have different numerators and denominators. In this sense, fractions are congruent if and only if their numerators and denominators are a constant multiple of each other.

In geometry, two figures are congruent if they have the exact same size and shape. The figures may be translated, rotated or even reflected. [Note: Some people distinguish a reflection by saying that it has a different orientation.] If two figures are congruent, they have the same corresponding lengths and angles, which might give us more information about the figures. Conversely, if we show that two figures have the same corresponding lengths and angles, then we know that they are congruent figures.

In the case of triangles, if two triangles have corresponding lengths (but we do not know anything about their angles), we can apply the Cosine Rule to conclude that they have the same corresponding angles. In fact, apart from SSS (side-side-side), we also have the methods of SAS (side-angle-side), ASA (angle-side-angle), and AAS (angle-angle-side) to prove triangles are congruent. Remember that SSA (side-side-angle) doesn't necessarily imply congruency, as in the Ambiguous case of Sine Rule.

Worked Examples

1. In isosceles triangle ABCABC with AB=BCAB=BC, let DD be the foot of the perpendicular from B B to ACAC. Show that ABD ABD and CBDCBD are congruent triangles.

Solution: Since the triangle is isosceles, we know that AB=BC AB = BC and BAD=BCD \angle BAD = \angle BCD . Then, AD2=AB2BD2=CB2BD2=CD2 AD^2 = AB^2 - BD^2 = CB^2 - BD^2 = CD^2 which gives us AD=CDAD = CD . Hence, by SAS, we get that triangles ABDABD and CBDCBD are congruent.


2. Prove that if two triangles have equal corresponding side lengths, then they are congruent.

Solution: Let ABCABC and DEFDEF be triangles such that


From the Cosine Rule, we know that cosABC=AB2+BC2CA22AB×BC=DE2+EF2FD22DE×EF=cosDEF. \cos \angle ABC = \frac{ AB^2 + BC^2 - CA^2}{2 AB \times BC } = \frac{ DE^2 + EF^2 - FD^2} { 2 DE \times EF} = \cos \angle DEF. Since angles in a triangle are in the range 0 0 ^ \circ to 180 180^\circ , it follows that ABC=DEF \angle ABC = \angle DEF . Similar equations hold for the other angles, which show that the corresponding angles are equal. Hence, the triangles are congruent.


3. ABCDABCD is a parallelogram. Let ACAC and BDBD intersect at EE. Show that triangles EABEAB and ECDECD are congruent.

Solution: Since ABCDABCD is a parallelogram, it follows that EAB=ECD \angle EAB = \angle ECD and DBA=EDC \angle DBA = \angle EDC . By vertical angles, AEB=CED \angle AEB = \angle CED and by AAA (angle-angle-angle), the two triangles are similar.

To show the triangles are congruent, it remains to show that two corresponding sides of the triangle have the same length. This follows immediately since ABCDABCD is a parallelogram, implying AB=CDAB = CD .

Note by Arron Kau
7 years, 3 months ago

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