Angles and Triangles ConstructionKonstruksi Sudut dan Segitiga

Construct ∠PQR which is equal to ∠BACDuplicating Angles

Steps:1. Construct a line QR (buatlah garis QR)2. In ∠BAC construct a circular arc with A as the

center such that it intersects AB at D and AC at E (pada ∠BAC buatlah busur dengan pusat A sehingga busur tersebut memotong AB di D dan AC di E)

3. With AD as the radius, construct an arc with Q as the center and intersects QR at S. (dengan jari-jari AD buatlah busur yang berpusat di Q dan memotong QR di S)

4. With DE as the radius construct an arc centered at S and intersect the arc that maked in step 3 at point T.

5. Construct a line passing through Q and T and named it QP.

6. The construction of ∠PQR which has the same measure ∠ABC is done.

Bisect ∠RPQBisecting Angles

Steps:1. Construct a circular arc centerd at P such that it

intersect PQ at S and PR at T (buatlah busur yang berpusat di P sehingga memotong PQ di S dan PR di T)

2. With S and T as the center points, construct two circular arcs (with equal radius) whics intersect at point U(dengan S dan T sebagai pusatnya, buatlah busur (jari-jari sama) dan berpotongan di titik U)

3. Join the points P and U such that PU is the line segment which bisects ∠RPQ so that ∠RPU = ∠UPQ (hubungkan titik P dan U sehingga PU adalah ruas garis yang membagi ∠RPQ sehingga ∠RPU = ∠UPQ )

Constructing a 90º Angle

Step 1:  Draw the arm AB.Step 2:  Place the point of the compass at B and draw

an arc that passes through A and interect the extension of AB at C.

Step 3:  With point A and C as the center, construct two arcs (radius greater than AB) so that intersect at point D.

Step 4:  Join the points B and D, now you have ∠ABD = 90o

Constructing a 45º Angle

We know that: 45 = ½ . 90So, to construct an angle of 45º, first construct a 90º angle and then bisect it.

Step 1:  Construct ∠BAC = 90oStep 2:  Construct the bisector of ∠BAC, and call it AR, this way we have ∠BAR = 45o

Constructing a 60º Angle

Step 1:  Draw the arm PQ.Step 2:  Place the point of the compass at P and draw

an arc that passes through Q.Step 3:  Place the point of the compass at Q and draw an

arc that passes through P so that intersect the arc in step 2 at point R

Step 4:  Join the points P and R, now you have ∠RPQ = 60o

Constructing a 30º Angle

We know that: 30 = ½ . 60So, to construct an angle of 30º, first construct a 60º angle and then bisect it.

Step 1:  Construct ∠BAC = 60oStep 2:  Construct the bisector of ∠BAC, and call it AR, this way we have ∠BAR = 30o

Constructing a 150º Angle

Construct ∠ABC = 150º

We know that: 150 = 90 + 60So, to construct an angle of 150º, first construct a 90º angle and then add with a 60o angle.

Step 1:  Construct ∠ABP = 90oStep 2:  Construct ∠PBC = 60o

Constructing Special Lines of TrianglesThere are 4 special lines in a triangle.1. Altitude (garis tinggi)2. Angle Bisector (garis bagi)3. Median (garis berat)4. Perpendicular Bisector (garis

sumbu)

Constructing a Perpendicular

a. Constructing a Perpendicular to a line segment through a point outside the line segment.(Menggambar garis tegak lurus terhadap suatu ruas garis melalui suatu titik di luar ruas garis)

Construct a perpendicular to line k passing through P.

STEPS:1. construct a circular arc centered at P so that

intersect line k at A and B2. Make two circular arcs having same

radius,centered at A and B, so that intersect at point C

3. Join the points P and C, so PC perpendicular to k, or PC⊥ k

Constructing a Perpendicular

b. Constructing a Perpendicular to a line segment through a point located on the line segment.(Menggambar garis tegak lurus terhadap suatu ruas garis melalui suatu titik yang terletak di ruas garis tersebut)

Construct a perpendicular to line k passing through M.

STEPS:1. construct a circular arc centered at M so that

intersect line k at A and B2. Make two circular arcs having same radius (radius

should be greater than MA),centered at A and B, so that intersect at point C

3. Join the points M and C, so MC perpendicular to k, or MC⊥ k

A B

C

D

EF

Altitude of Triangle

DC, EA, FB are the altitudesof triangle ABC

DC, EA and FB intersect at single point.

An altitude of a triangle is a line passing through a vertex of the triangle and perpendicular to opposite side.

T

Constructing Altitude of Triangle

A B

C

On the triangle ABC on the left, construct an altitude through the point B

Step 1:  with B as the center point, construct a circular arc which intersect AC at P and Q

Step 2:  At P and Q as the center points, construct two arcs with same radius which intersect at point R

Step 3:  joint the point B and R to construct a line which intersect AB at S

Step 4: the line BS is the altitude through point B.

Angle Bisectors in a Triangle(Garis Bagi Segitiga)

DC, EA, FB are angle bisectorsof triangle ABC

DC, EA and FB intersect at single point.

An angle bisector of a triangle is a line passing through a vertex of the triangle which bisect (divides into two equal parts) the corresponding angles.

A B

C

D

EF

Z●● **

Constructing Angle Bisectorin a Triangle

A B

C

On the triangle ABC on the left, construct an angle bisector through the

point B

Step 1:  with B as the center point, construct a circular arc which intersect AB at P and BC at Q

Step 2:  At P and Q as the center points, construct two arcs with same radius which intersect at point R

Step 3:  joint the point B and R to construct a line which intersect AC at T

Step 4: the line BT is the angle bisector through point B.

Perpendicular Bisectors of a Triangle (Garis Sumbu)

A perpendicular bisector of a triangle is a line passing through the midpoint of a side and perpendicular to the side

A B

C

Constructing a PerpendicularBisector of a line Segment

Construct a perpendicular bisectors of line segment AB

STEPS:1. construct a circular arc centered at A so that

intersect segment AB2. With the same radius at step one, construct a

circular arc centered at B so that intersect the arc that made in step one in point P and Q

3. Join the points P and Q, so PC perpendicular to AB, or PQ⊥ AB

Median of a Triangle(Garis Berat)

A median of a triangle is a line passing through a vertex

and the midpoint of the opposite side.

A B

C

D

EF

DC, EA, FB are mediansof triangle ABC

DC, EA and FB intersect at single point.

Constructing Medianin a Triangle

P Q

R

On the triangle PQR on the left, construct a median through the point P

Step 1:  construct a perpendicular bisector of side RQ which intersect RQ at point S

Step 2: Join point P and SStep 3:  the segment PS is the median through point P