The Angle Sum of a Triangle
Prove that x + y + z = 180°
A
B
C
x
y
z
Construct PQ through A so that PQ||BC
A
B
C
Q
P
x
y
z
∠PAB = y° (alternate angles, PQ||BC)
A
B
C
Q
P
x
y
z
y
∠QAC = z° (alternate angles, PQ||BC)
A
B
C
Q
P
x
y
z
z y
Now PAQ is a straight line
A
B
C
Q
P
x
y
z
z y
x + y + z = 180° (PAQ is a straight line)
A
B
C
Q
P
x
y
z
z y
x + y + z = 180°
A
B
C
x
y
z
The angle sum of a triangle is 180°
A
B
C
x
y
z
A different proof for the same result
Prove that x + y + z = 180°
A
B
C
x
y
z
Produce BC to P A
B
C P
x
y
z
At C construct CQ||BA A
B
C
Q
P
x
y
z
∠ACQ = x° (alternate angles, PQ||BC)
A
B
C
Q
P
x
y
z x
∠PCQ = y° (corresponding angles, PQ||BC)
A
B
C
Q
P
x
y
z y x
Now BCP is a straight line A
B
C
Q
P
x
y
z y x
A
B
C
Q
P
x
y
z y x
x + y + z = 180° (BCP is a straight line)
x + y + z = 180° A
B
C
x
y
z
The angle sum of a triangle is 180°
A
B
C
x
y
z
Top Related