ART Liek Wilardjo From isosceles triangle Full text
FROM /SO.W
ＧﾣＱｾＭＬ＠
TR/AN(;'LE TO NOREf lAURA TES COMPTOl\:', DE BROG'LJE,
l>A V/.\:\'ON, AND THOMSON
Ltek W!lardjo
FROM ISOSCELES TRIANOLE TO NOBEL LAURATES COMPTON, IJE
BR06'LIE, IJA VISSON, AND THOMSON
Liek Wilardjo
Program Studi Teknik Elektro
Fakultns Teknik Elektronika dan Komputer
OKSW
Jalan Diponegoro 52-60. Salatiga 50711
1. Pengantar
A1tikel di bawah ini berisi tentang planimetri elementer. yakni segitiga samakaki dan datil
p,·thagoras padn segitiga siku-siku. Dalam artikel ini ditampi!kan contoh sederh::mn
ｴ･ｮｾｭｧ＠
geometri Euklides-an. Kedua jenis segitiga itu dan sifatnya dipakai untuk menunjukkan
pembuktian sifat (propcrttes) atau datil (propositions) secara matematis.
Melalui planimetri. Be11rand Russell sewaktu masih bocah pertama kah
matematika dari guru priYatnya0 l Melatui dalil ｐｾｴｨ｡ｧｯｲｳＮ＠
ｭ･ｰｬｾｪ｡ｲｩ＠
Sokrates "memperagakan" bahwa
matematika (menurut dia) ialah kebenaran abadi. Seorang bocat1. anak budak belian. mampu
menemukan sendiri bukti datil PythagorasUJ Demonstrasi Sokrates itu menandai lahirnya
discovet:l' atau inquit)' method yang sampai sekarang masih dipakai di bidang pendidikan.
Anamnesis yang dilakukan dokter terhadap pasiennya juga didasarkan pada keyakinan yang
sam a.
Datil Pythagoras dipakru Einstein untuk mengungkapkru1 kekararan (invariance) skalar
berupa kuadrat Yektor-4 pusa-tenaga (momentum-energytour-vector). yru1g "bermuara" pad a
"rumus"-nya yang terkenal. yakni E
rm_.J._ Louis Victor de Broglie kemudian menarik
analogi dari cara Einstein itu. selungga ia menemukan keseduaan zarat1-gelombang (particle-
wave duality). dengru1 persrunarumya p = h A.= flk. Dengan penemurumya itu Louis de Broglie
bukru1 hanya memperoleh gelar doktor dari Umversite de Paris (dengru1 Profesor Paul
Lru1geyin sebagai promotornyal 3> tetapi juga dianugerahi hadiail Nobel dalrun fisika.
Hasil yang diperoleh Einstein. de Broglie. dru1 sebelumnyajuga oleh Max Planck (E
hv
ftw. untuk foton) ternyata mengilhruni kru·ya-k:arya keilmuru1 lain. yang juga "menyabet"
hadiah Nobel. Jadi dengan artikel ini say a hendak menunjukkru1 batm·a kebenaran 'ang
tampaknya sederhru1a terny ata bisa membenih (seminal)
125
2. Isosceles Triangle
c
ABC is a triangle in "·hich AH is the base
and C the apex lfthe base angles. LA and
L
B. are equal. then side A(' is equal to side
R(' .
and thus AI:K IS an ISosceles tnangte.
Proof:
( I)
ｩＮ｟ＡＺ［ｾｬｵＬ＠
ｌｾ｟ＮＺａＬ｜＠
A
D
B
Dra" the tnangle in a piece of clear I
transparent plastic Draw the bisector of the apex angle L C until it mtercepts base AB
at a point. D. It is obYious that CD is perpendicular to AB. No\Y fold the paper along
CD. You \Yill see that A and B comcide and the right triangles ACD and BCD m erlap
perfectly. "proYing" that AC is indeed equal to BC.
What "e hm e just done is actually not praYing that ABC is an isosceles triangle. It is a
"platonic"" ay of demonstrating experimentally or obsen ationally that AC and B('
are of equal length. To Plato. geometry is real. i.e.. it deals \Yith concrete entities.
(2)
To proYe the theorem mathematically.
c
assume that AC i BC _say. that AC' is
longer than BC . Then there is a point.
call it E. on AC such that B(' =EC.
Draw line EF \\ith the point F on side
B
RC _angle ,:::.. CEF being equal to angle ._:
CBE. Then the triangles CEF and CBE
identical in
shape. two. and consequently all three. of their corresponding angles being equal.
Therefore there is a proportionality bet\yeen the corresponding sides of the two
----= EFx BE= FC x C'B
Ｍｾ
trianglesfi.e.
CE: EF: FC = CB: BE: EC. or CE x EC
I
This is a three-side equation with the left the middle and the right sides. From the
equalit: of the left E BROGLIE,
l>A V/SSON, ANJ> THOMSON
Liek Wilurdju
The conclusion that £('
= F('
leads to an absurd situation. Thus the opposite of our
assumption. i.e .. that A(' = B('. is true by the logic of "rcductw ud ahsurdum''. {!uod
erat demosntrcmdum (Q.E.D.)
3 Pythagorean Theorem
ABC Is a nght-angle tnangle. The
1\hile RC the
ｨｾ＠
A IS')()'' AH and A(· are the nght sides.
｡ｮｧｬ･ｾ＠
potenuse. Calling AB. A(·. and B( · c. h. and u. respectiYely.
Pythagorean theorem states that a
2
= h-) + c 2 . I.e .. the square of the hypotenuse IS.
equal to the swn of the squares of the right
ｳｩ､･ｾ＠
Proof'
A
( 1)
R
For an isosceles. right triangle. h =c. so
''e ha' e to proYe that
a2
=b2 + c2 = 2 h2
(or a= h.fi
c
DrmY three more right triangles of the
same shapes and size to that of triangle
ABC. so that ''e haw a square. named
BCDE. \\hose centre is A The area of
square BCDE is L 1 = a 2 . a being its side
E
The area of triangle ABC is
L2 ={he== *h 2 . since h = c
From the
figure it is
obYJOUS that L1 = 4L 2 . therefore a 2
= 4 x 1h 2 = 2h 2 . giYing a= h.fi.
thus proYing
the theorem for the special case of an isosceles. right-angle triangle
127
Techne Jurnalllmtah Elcktrotekntb.a VoL') No 2 01-.tobct 21J l 0 Hal ＱＲｾ＠
According to Plato (in "Dialogue
vFith
ll!
Meno") 1 tlus proof was taken by Socrates as a
supporting eYidence of the truth of his belief that mathematics is eternal truth. In front of
his friend. Meno. Socrates demonstrated that a boy "ho was the son of a stare could
rediscoYer the Pythagorean theorem by himself: Socrates merely guiding him with a
series of questions This is "·hat "e call the discoYery or the inquiry method of teaching.
{'
( 2)
For Ｚｾｮｙ＠
nght-angle triangle. the proof
goes as follows
2
Consider the right-angle
triangle ABC with the right angle at the
A
Yertex C. sub tended bet\\ een the right
B
D
a and CA = h and the
sides
hypotenuse ｾＮＺ＠
as 1ts base.
AB Name the base angles a and jJ. at \·erte:-.. A
and \ ene-...: B.
respecth·ely. The right-angle triangle ABC is defined. i.e .. it is fully-specified. when its
base and either one of its base angles is gh en. Thus its area L is completely determined
by c and jJ. As angles or functions of angles can be considered as dimensionless. I. must
be proportional to the square of the base. c. times some fimction of a base angle. say jJ
Dra"
C'D with[) on AB and CD
j_
AB. Since the sum of the three i11terior angles in
any triangle is equal to 180°. the angle yat the Yertex C of the right-angle triangle ADC
must be equal to the angle j3 at the Yertex B of the right-angle triangle ACB. Then the
2
area oftriangle ADC L1. is proportional to h f(y)
CDB.
2
is proponional to a f(fJ) But from the ｦｩｾＬＧｬ･＠
h 2f
ＧﾣＱｾＭＬ＠
TR/AN(;'LE TO NOREf lAURA TES COMPTOl\:', DE BROG'LJE,
l>A V/.\:\'ON, AND THOMSON
Ltek W!lardjo
FROM ISOSCELES TRIANOLE TO NOBEL LAURATES COMPTON, IJE
BR06'LIE, IJA VISSON, AND THOMSON
Liek Wilardjo
Program Studi Teknik Elektro
Fakultns Teknik Elektronika dan Komputer
OKSW
Jalan Diponegoro 52-60. Salatiga 50711
1. Pengantar
A1tikel di bawah ini berisi tentang planimetri elementer. yakni segitiga samakaki dan datil
p,·thagoras padn segitiga siku-siku. Dalam artikel ini ditampi!kan contoh sederh::mn
ｴ･ｮｾｭｧ＠
geometri Euklides-an. Kedua jenis segitiga itu dan sifatnya dipakai untuk menunjukkan
pembuktian sifat (propcrttes) atau datil (propositions) secara matematis.
Melalui planimetri. Be11rand Russell sewaktu masih bocah pertama kah
matematika dari guru priYatnya0 l Melatui dalil ｐｾｴｨ｡ｧｯｲｳＮ＠
ｭ･ｰｬｾｪ｡ｲｩ＠
Sokrates "memperagakan" bahwa
matematika (menurut dia) ialah kebenaran abadi. Seorang bocat1. anak budak belian. mampu
menemukan sendiri bukti datil PythagorasUJ Demonstrasi Sokrates itu menandai lahirnya
discovet:l' atau inquit)' method yang sampai sekarang masih dipakai di bidang pendidikan.
Anamnesis yang dilakukan dokter terhadap pasiennya juga didasarkan pada keyakinan yang
sam a.
Datil Pythagoras dipakru Einstein untuk mengungkapkru1 kekararan (invariance) skalar
berupa kuadrat Yektor-4 pusa-tenaga (momentum-energytour-vector). yru1g "bermuara" pad a
"rumus"-nya yang terkenal. yakni E
rm_.J._ Louis Victor de Broglie kemudian menarik
analogi dari cara Einstein itu. selungga ia menemukan keseduaan zarat1-gelombang (particle-
wave duality). dengru1 persrunarumya p = h A.= flk. Dengan penemurumya itu Louis de Broglie
bukru1 hanya memperoleh gelar doktor dari Umversite de Paris (dengru1 Profesor Paul
Lru1geyin sebagai promotornyal 3> tetapi juga dianugerahi hadiail Nobel dalrun fisika.
Hasil yang diperoleh Einstein. de Broglie. dru1 sebelumnyajuga oleh Max Planck (E
hv
ftw. untuk foton) ternyata mengilhruni kru·ya-k:arya keilmuru1 lain. yang juga "menyabet"
hadiah Nobel. Jadi dengan artikel ini say a hendak menunjukkru1 batm·a kebenaran 'ang
tampaknya sederhru1a terny ata bisa membenih (seminal)
125
2. Isosceles Triangle
c
ABC is a triangle in "·hich AH is the base
and C the apex lfthe base angles. LA and
L
B. are equal. then side A(' is equal to side
R(' .
and thus AI:K IS an ISosceles tnangte.
Proof:
( I)
ｩＮ｟ＡＺ［ｾｬｵＬ＠
ｌｾ｟ＮＺａＬ｜＠
A
D
B
Dra" the tnangle in a piece of clear I
transparent plastic Draw the bisector of the apex angle L C until it mtercepts base AB
at a point. D. It is obYious that CD is perpendicular to AB. No\Y fold the paper along
CD. You \Yill see that A and B comcide and the right triangles ACD and BCD m erlap
perfectly. "proYing" that AC is indeed equal to BC.
What "e hm e just done is actually not praYing that ABC is an isosceles triangle. It is a
"platonic"" ay of demonstrating experimentally or obsen ationally that AC and B('
are of equal length. To Plato. geometry is real. i.e.. it deals \Yith concrete entities.
(2)
To proYe the theorem mathematically.
c
assume that AC i BC _say. that AC' is
longer than BC . Then there is a point.
call it E. on AC such that B(' =EC.
Draw line EF \\ith the point F on side
B
RC _angle ,:::.. CEF being equal to angle ._:
CBE. Then the triangles CEF and CBE
identical in
shape. two. and consequently all three. of their corresponding angles being equal.
Therefore there is a proportionality bet\yeen the corresponding sides of the two
----= EFx BE= FC x C'B
Ｍｾ
trianglesfi.e.
CE: EF: FC = CB: BE: EC. or CE x EC
I
This is a three-side equation with the left the middle and the right sides. From the
equalit: of the left E BROGLIE,
l>A V/SSON, ANJ> THOMSON
Liek Wilurdju
The conclusion that £('
= F('
leads to an absurd situation. Thus the opposite of our
assumption. i.e .. that A(' = B('. is true by the logic of "rcductw ud ahsurdum''. {!uod
erat demosntrcmdum (Q.E.D.)
3 Pythagorean Theorem
ABC Is a nght-angle tnangle. The
1\hile RC the
ｨｾ＠
A IS')()'' AH and A(· are the nght sides.
｡ｮｧｬ･ｾ＠
potenuse. Calling AB. A(·. and B( · c. h. and u. respectiYely.
Pythagorean theorem states that a
2
= h-) + c 2 . I.e .. the square of the hypotenuse IS.
equal to the swn of the squares of the right
ｳｩ､･ｾ＠
Proof'
A
( 1)
R
For an isosceles. right triangle. h =c. so
''e ha' e to proYe that
a2
=b2 + c2 = 2 h2
(or a= h.fi
c
DrmY three more right triangles of the
same shapes and size to that of triangle
ABC. so that ''e haw a square. named
BCDE. \\hose centre is A The area of
square BCDE is L 1 = a 2 . a being its side
E
The area of triangle ABC is
L2 ={he== *h 2 . since h = c
From the
figure it is
obYJOUS that L1 = 4L 2 . therefore a 2
= 4 x 1h 2 = 2h 2 . giYing a= h.fi.
thus proYing
the theorem for the special case of an isosceles. right-angle triangle
127
Techne Jurnalllmtah Elcktrotekntb.a VoL') No 2 01-.tobct 21J l 0 Hal ＱＲｾ＠
According to Plato (in "Dialogue
vFith
ll!
Meno") 1 tlus proof was taken by Socrates as a
supporting eYidence of the truth of his belief that mathematics is eternal truth. In front of
his friend. Meno. Socrates demonstrated that a boy "ho was the son of a stare could
rediscoYer the Pythagorean theorem by himself: Socrates merely guiding him with a
series of questions This is "·hat "e call the discoYery or the inquiry method of teaching.
{'
( 2)
For Ｚｾｮｙ＠
nght-angle triangle. the proof
goes as follows
2
Consider the right-angle
triangle ABC with the right angle at the
A
Yertex C. sub tended bet\\ een the right
B
D
a and CA = h and the
sides
hypotenuse ｾＮＺ＠
as 1ts base.
AB Name the base angles a and jJ. at \·erte:-.. A
and \ ene-...: B.
respecth·ely. The right-angle triangle ABC is defined. i.e .. it is fully-specified. when its
base and either one of its base angles is gh en. Thus its area L is completely determined
by c and jJ. As angles or functions of angles can be considered as dimensionless. I. must
be proportional to the square of the base. c. times some fimction of a base angle. say jJ
Dra"
C'D with[) on AB and CD
j_
AB. Since the sum of the three i11terior angles in
any triangle is equal to 180°. the angle yat the Yertex C of the right-angle triangle ADC
must be equal to the angle j3 at the Yertex B of the right-angle triangle ACB. Then the
2
area oftriangle ADC L1. is proportional to h f(y)
CDB.
2
is proponional to a f(fJ) But from the ｦｩｾＬＧｬ･＠
h 2f