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Thread: Similarity of triangle

  1. #1
    Jul 2005

    Similarity of triangle

    I have noticed that there are many excercises which are easy to solve if you know the concept as similarity of triangle (sometimes only way to solve), so look at the attachment.
    How is it possible that KB/DC=AB/AC
    I dont know how to write these equations.
    Attached Thumbnails Attached Thumbnails Similarity of triangle-wt.jpg  
    Last edited by totalnewbie; Sep 18th 2005 at 04:21 AM.
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  2. #2
    Junior Member
    Aug 2005

    Ok, the concept of similar triangles is VERY useful. It is a big help for you that you master this concept.

    How to do so ? You usually have 2 triangles. So, you rotate one so that both seem to be on the same point of view (figure 1).

    Now, we do that we your figure (figure 2). Ok, since you don't give any numerical question. I'll be general.

    First, to use the concept of similar triangles, you must prove they are similar (very important, never take it for granted if not said). For that, first identify the homologous sides and angles. It is easy when you have already put the figures in the same point of view. So angle D of the 1st triangle goes with angle K of the second triangle, C of the 1st with B of the 2nd and B of the 1st with A of the 2nd. For the sides, the side DB of the 1st triangle goes with KA of the 2nd triangle...

    Now tow triangles are similar when all the measure of the homologous angles are equal (here m(D)=m(K), m(C)=m(B)...) and their sides have proportional lengths (we can k the proportion). So k=m(DB)/m(KA)=m(DC)/m(KB)=m(CB)/m(BA) (so I took the quotient of the homologous sides. It is very important. Making sure that the sides on the numerator are all of the same triangle and those at the denominator are of the other triangle.) Why take care of these details ? Example : if triangle DCB was twice as big as triangle KBA. Their angles would still be equal (since the sum is 180 degrees always it is impossible to have angles twice as big anyway) and every side of DCB is twice as big as its homologous side in KBA. Since the proportion (big triangle/small triangle) here is 2 (twice as big). So k=2=m(DB)/m(KA)=m(DC)/m(KB)=m(CB)/m(BA). If I did (small triangle/big triangle) which is m(KA)/m(DB)=m(KB)/m(DC)=m(BA)/m(BC)=1/2. It is all logic.

    Now, it is too long proving that every angle is equal to its homologous angle and every side has the same proportion to its homologous side so we have shortcuts. It suffies to prove only one of these asserts to conclude that 2 triangles are similar :

    1) AA. You have 2 homolougous angles which are equal (example : you have m(D)=m(K) and m(B)=m(A)).
    2) SSS. All homologous sides are proportional with the same proportion. So here you would prove m(DB)/m(KA)=m(DC)/m(KB)=m(CB)/m(BA). (So this is useful when you have no idea about the angles).
    3) SAS. you have 2 homologous sides with same proportion and the angle between them (not any angle !!!) is equal to its homolougous angle (example : m(DB)/m(KA)=m(DC)/m(KB) and m(D)=m(K)).

    Any one of these conditions proves that your 2 triangles are similar.

    What for ? Proving one of these asserts gives you the possibility to use all the other properties of the similar triangles. Example: you have two triangles and you only know that they both have two angles of 50 degrees and 20 degrees. This info is sufficient to conclude that the two triangles are similar. So all the sides are proportional with the same proportion and the other angle of the 1st triangle (180-50-20=110) is equal to the other in the 2nd triangle.

    So it can be very useful. I give a very used application (figure 3) : here the 2 triangle are already put in the same point of you so the homologous sides and angles are easy to see. (first step done)

    We are looking for the measure of DB.

    Ok, you have the case AA (m(D)=m(K)=80' and m(C)=m(B)=10') So the 2 triangles are similar. (second step done)

    So the fact that they are similar tells you that their homologous angles are equal and there exists a proportionallity between their homologous sides. k=m(DB)/m(KA)=m(DC)/m(KB)=m(CB)/m(BA). We know one of the quotients : m(CB)/m(BA) = 2/1=2 so k=2=m(DB)/m(KA)=m(DC)/m(KB). So you can find DB easily since we know 2=m(DB)/m(KA)=m(DC)/m(KB) so 2=m(DB)/m(KA) and 2=m(DC)/m(KB) but we just need here
    so 2m(KA)=m(DB) and we know m(KA)=0.5 so m(DB)=1

    Done !
    Attached Thumbnails Attached Thumbnails Similarity of triangle-fig1.gif   Similarity of triangle-fig2.gif   Similarity of triangle-fig3.gif  
    Last edited by hemza; Sep 18th 2005 at 09:32 AM.
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