Top Drawer Teachers: Two proofs of the angle sum of a triangle

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The Angle Sum of a
Triangle

Prove that x + y + z = 180°
A
x

B

y
z

C

Construct PQ through A
so that PQ||BC
P

A
x

B

Q

y
z

C

∠PAB = y°
(alternate angles, PQ||BC)
P

A
y x

B

Q

y
z

C

∠QAC = z°
(alternate angles, PQ||BC)
P

A
y x

B

z

Q

y
z

C

Now PAQ is a straight line
P

A
y x

B

z

Q

y
z

C

x + y + z = 180°
(PAQ is a straight line)
P

A
y x

B

z

Q

y
z

C

x + y + z = 180°
A
x

B

y
z

C

The angle sum of a triangle is 180°
A
x

B

y
z

C

A different proof for the
same result

Prove that x + y + z = 180°
A
x

B

y
z

C

Produce BC to P
A
x

B

y
z

C

P

At C construct CQ||BA
A
x

B

Q

y
z

C

P

∠ACQ = x°
(alternate angles, PQ||BC)
A

x

B

Q

y
z

x

C

P

∠PCQ = y°
(corresponding angles, PQ||BC)
A

x

B

Q

y
z

x

C

y

P

Now BCP is a straight line
A
x

B

Q

y
z

x

C

y

P

x + y + z = 180°
(BCP is a straight line)
A
x

B

Q

y
z

x

C

y

P

x + y + z = 180°
A
x

B

y
z

C

The angle sum of a triangle is 180°
A
x

B

y
z

C

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