Lesson3 - Similar triangles; The Proportion of Corresponding Sides (script)
teacher: Dragana Tepavac
class: VIII2
subject: mathematics
September 2014
Lesson 3: Similar triangles; The Proportion of Corresponding Sides
(Пропорционалност страница сличних троуглова)
HEIGHT
ВИСИНА
AREA
ПОВРШИНА
CORRESPONDING SIDES
ОДГОВАРАЈУЋЕ СТРАНИЦЕ
SCOPE
ОБИМ
QUOTIENT
КОЛИЧНИК
RATIO
РАЗМЕРА
IN THE SAME RATIO
У ИСТОЈ РАЗМЕРИ
PROPORTIONALITY COEFFICIENT
(PROPORTIONALITY FACTOR,
PROPORTIONALITY CONSTANT)
КОЕФИЦИЈЕНТ СЛИЧНОСТИ
(ПРОПОРЦИОНАЛНОСТИ)
If two triangles are similar, and their two angles are congruent then opposite sides of those angles
are CORRESPONDING SIDES
Knowing Thales theorem we can say:
Once the triangles are similar, the corresponding sides of those triangles are in proportion.
If triangle ABC ~ triangle DEF then:
𝐴𝐵
𝐷𝐸
=
𝐵𝐶
𝐸𝐹
=
𝐴𝐶
𝐷𝐹
= k (proportionality constant)
If a1, b1, c1 and a2, b2, c2 are corresponding sides of two similar triangles, k is proportionaly
constant then: a1 = k* a2 , b1 = k* b2 , c1 = k* c2
Example: Find DC
1.similarity of triangles
2. length of line segment DC using formula
proportionality constant
for
If two triangles are similar, the proportion of their scopes is = k
Knowing that triangles are similar we have corresponding sides a1, b1, c1 and a2, b2, c2 and
a1 = k* a2 , b1 = k* b2
O1 = a 1 + b 1 + c 1
;
, c1 = k* c2
O2= a2 + b2 + c2
Then O1/O2 = ...(in notes)......= k
Example: In triangle ABC a = 5 cm ; b = 6 cm ; c = 7 cm. Find the scope of triangle which is
similar to ABC if k= 3/2.
Practise: book, page 16; 10a, 11a, 12a
HOMEWORK book, page 16; 10bc, 11b, 12b
Theorems of similarity
median - тежишна дуж
angle bisector - симетрала угла
line segment bisector - симетрала странице
Suppose that we have two triangles ABC and A1B1C1
1. theorem: If two angles in triangle ABC are congruent with two angles in triangle A1B1C1 those two
triangles are similar ( ABC ~ A1B1C1 )
h1 , h2 - heights of triangle ABC
are triangles ABA1 and CBC1 similar ?
∢ AA1B = ∢ CC1B = 90°
∢ ABA1 = ∢ CBC1 (same angle for both triangles)
---> (1.theorem) ABA1 ~ CBC1
Knowing that we can prove that proportion of corresponding heights in two similar triangles equals
constant k.
We know that triangle ABC ~ triangle DEF
𝐴𝐵
𝐷𝐸
=
𝐵𝐶
𝐸𝐹
=
𝐴𝐶
𝐷𝐹
=k
hc - height of triangle ABC
hf - height of triangle DEF
can we prove that triangle CC1B ~ triangle FF1E ?
∢ CC1B = ∢ FF1E = 90°
∢ CBC1 ≅ ∢ FEF1 (angles with parallel sides) ---> (1.theorem) triangle CC1B ~ triangle FF1E
𝐵𝐶
Then hc : hf =
=k
𝐸𝐹
Suppose that P1 - area of A1B1C1 , and P2 - area of A2B2C2 . If triangle A1B1C1 ~ triangle A2B2C2 then
P1 : P2 = k2
(use proof with corresponding heights)
If
P1 =
𝑎1ℎ𝑐
2
and
P2 =
𝑎2ℎ𝑓
2
hc : hf =
k, triangles are similar then
then P1 : P2 =
𝑎1ℎ𝑐
2
:
𝑎2ℎ𝑓
2
𝑎1
𝑎2
=
𝑏1
𝑏2
=
𝑐1
𝑐2
= a1hc : a2hf = k⋅ k = k2
Example: In right triangle a = 5 cm, b = 12 cm. FInd a1, b1, c1 of similar triangle if P1 = 270 cm2.
2.theorem: If two sides of triangle ABC are in proportion with two sides in triangle A1B1C1, also
correponding angles are congruent, then those triangles are similar.
=k
Example:
Triangles, ABC, EFG
s1, s2 - angles bisectors of ∢EFG and
∢FGE
∢ between s1, s2 is 119°
are triangles ABC and EFG similar?
Suppose that ∢FGE = 2𝛽, and
∢EFG = 2𝛼
In triangle OFG 𝛼 + 𝛽 + 119° = 180° --> 𝛼 + 𝛽 = 61° ----> ∢FGE + ∢EFG = 122°
From triangle EFG we have ∢FGE = 180° - 122° = 58°
2
AC : EG = AB : DE = k because 2 : 2 = 3 : 4
3
Angles between given sides equal 58° ---> triangles ABC and EFG are similar.
Example: we can prove that proportion of corresponding medians in to similar triangles equals
constant k.
tc - median of triangle ABC
tf - median of triangle DEF
(median divides opposite side into two
equal parts)
1
1
2
2
AC1 : DF1 = AB : DE = AB : DE = k
AC : DF = k
∢ C1AC ≅ ∢ FDF1 = 𝛼 ----> (2.theorem) triangles C1AC and DFF1 are similar
---> tc : tf = CC1 : FF1= k
3. theorem: If sides of triangle ABC are in proportion with correponding sides of triangle A1B1C1 those
two triangles are similar. (often used with isosceles triangles)
Example:
Are triangles ABC and BCD similar?
ABC - isosceles triangle
BCD - isosceles triangle
1
4
3
3
2:1 =2:
3:2=
3
=
6
4
=
3
2
---> sides are in proportion so triangles are
2
similar (3.theorem)
4. theorem: If two sides in one triangle are inproportion with corresponding sides in other triangle,
and angles opposite longer sides are congruant, those two triangles are similar.
Example: are triangles ABC and DEF similar?
2,4 : 3 =
4: 5 =
4
4
5
5
∢A = ∢D = 90° ----> triangles are similar
Book, page 17: 15, 17, 18, 19