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This would be alot easier to answer if I could draw a picture, but we can do it without.
Say you have two triangles -- ABC (smaller) and DEF (larger) -- with proportional sides. That is, AB/DE = BC/EF = CA/FD. How do we prove that the two triangles are similar - that is, that the angles are equal?
Measure down from the top angle (angle E) of the larger triangle the distance of the smaller triangle's proportional side. The do the same the in other direction from the top angle, using the smaller triangle's corresponding proportional side. Connect the two points and you will have effective overlaid the smaller triangle on the larger lined up at the top angles (angle B and angle E).
The base of the smaller triangle will now be parallel to the base of the larger. This creates a situation where two parallel lines are crossing transversals. Therefore, angle C of ABC and angle F of DEF are corresponding angles and are equal. Similarly, angle A of ABC and angle D of DEF are corresponding and equal.
You know that if two angles of the two triangles are equal, then the third angle must also be equal.
This proves that the triangles are similar -- they have proportional sides and their angles are equal.
Proof of S S S similarity:
Two triangles are similar if for each angle of one triangle , there is an equal angle in another triangle. And they are the corresponding angles .
The side opposite to the corresponding angles are called corresponding sides in the two triangles.
The corresponding sides of two similar triangles bear the same ratio.
Proof: Let ABC and A'B'C' be two similar triangles. Since they are similar, we can assume that angles A=A', B = B' and C=C'.Let us assume again without loss of generality that ABC is the bigger triangle.
Construction: With compass take A'B' as radius and A as centre mark B1 on AB where AB1 = A'B' . Simalilarly mark C1 on AC such that AC1 = A'C'
Now the triangles AB1C1 and A'B'C' are congruent as
AB1=A'B' and AC1 =A'C' by construction. Angle A=A' being equal angles of similar triangles. So, SAS postulate holds good for congruency.
Therefore, the angles AB1C1 = B' = B and angle AC1B1 =C' = C
This confirms that the line B1C1 is parallel to the line or side BC.
The line B1C1 is parallel to BC, B1 being on AB and C1 beeing AC , the ratio, AB1/ AB = AC1/AC = B1C1/BC by Thales intercept theorem.
The converse is also true. That is, if the corresponding sides of two triangles are in the same ratio, then the triangles are similar or the corresponding angle are equal. The proof involves, identifying the smaller triangle A'B'C' and AB1C1 and applying the Thales intercept theorem and prove that B1C1 is parallel as AB1/AC= AC1/AC. Thus the Parallel property of B1C1 with BC gives us the corresponding angles AB1C1 = B =B'and AC1B1 = C=C'. Consequently the remaining angle A has to be equal to A'.
Is this proof helpful?
to use SSS to find the similarity just divide the corresponding sides by each other, and see if all of the get the same answer
for ex: to prove that triangle ABC = triangle DEF and lets say AB = DE
DC= EF and CA = FD just divide a/d b/e c/f and if they are all the same then the triangle are similar.
Ok, so we just covered this, and both of the previous answers are incorrect. You cannot use SSS to prove similarities in triangles (you can however use AAA). If all three sides are the same, then of course the triangles will be congruent. Not similar.
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