Proof: Triangle And Line Parallel

Let's prove the following theorem:

if distance WY = distance XY and ray YP bisects ∠XYZ and m∠WYZ = 180, then WX || YP

Proof:

View as a tree | View dependent proofs | Try proving it

Given

1	distance WY = distance XY
2	ray YP bisects ∠XYZ
3	m∠WYZ = 180

Proof Table

#	Claim	Reason
1	m∠XYP = m∠PYZ	if ray YP bisects ∠XYZ, then m∠XYP = m∠PYZ (Angle Bisector Property)
2	point P lies in interior of ∠XYZ	if ray YP bisects ∠XYZ, then point P lies in interior of ∠XYZ (Bisector of an Angle is in the Interior)
3	(m∠XYP) + (m∠PYZ) = m∠XYZ	if point P lies in interior of ∠XYZ, then (m∠XYP) + (m∠PYZ) = m∠XYZ (Angle Addition Theorem 2)
4	m∠XWY = m∠WXY	if distance WY = distance XY, then m∠XWY = m∠WXY (Angles of an Isosceles Triangle)
5	(m∠XWY) + (m∠WXY) = m∠XYZ	if m∠WYZ = 180, then (m∠XWY) + (m∠WXY) = m∠XYZ (Exterior Angle Equal to Sum of Nonadjacent 2)
6	(m∠WXY) + (m∠WXY) = m∠XYZ	if (m∠XWY) + (m∠WXY) = m∠XYZ and m∠XWY = m∠WXY, then (m∠WXY) + (m∠WXY) = m∠XYZ (Substitution 8)
7	m∠WXY = (m∠XYZ) ⋅ (1 / 2)	if (m∠WXY) + (m∠WXY) = m∠XYZ, then m∠WXY = (m∠XYZ) ⋅ (1 / 2) (Double to Half)
8	(m∠XYP) + (m∠XYP) = m∠XYZ	if (m∠XYP) + (m∠PYZ) = m∠XYZ and m∠XYP = m∠PYZ, then (m∠XYP) + (m∠XYP) = m∠XYZ (Substitute 2)
9	m∠XYP = (m∠XYZ) ⋅ (1 / 2)	if (m∠XYP) + (m∠XYP) = m∠XYZ, then m∠XYP = (m∠XYZ) ⋅ (1 / 2) (Double to Half)
10	m∠WXY = m∠XYP	if m∠WXY = (m∠XYZ) ⋅ (1 / 2) and m∠XYP = (m∠XYZ) ⋅ (1 / 2), then m∠WXY = m∠XYP (Transitive Property of Equality Variation 1)
11	WX \|\| YP	if m∠WXY = m∠XYP, then WX \|\| YP (Alternate Interior Angles Theorem 4)

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