Proof: Similar Triangles Example 3
Let's prove the following theorem:
if ∠ZXY is a right angle and ∠XPY is a right angle and m∠YPZ = 180, then △PXZ ∼ △XYZ
Proof:
Given
| 1 | ∠ZXY is a right angle |
|---|---|
| 2 | ∠XPY is a right angle |
| 3 | m∠YPZ = 180 |
| # | Claim | Reason |
|---|---|---|
| 1 | ∠YPX and ∠XPZ are supplementary | if m∠YPZ = 180, then ∠YPX and ∠XPZ are supplementary |
| 2 | (m∠YPX) + (m∠XPZ) = 180 | if ∠YPX and ∠XPZ are supplementary, then (m∠YPX) + (m∠XPZ) = 180 |
| 3 | m∠YPX = 90 | if ∠XPY is a right angle, then m∠YPX = 90 |
| 4 | 90 + (m∠XPZ) = 180 | if (m∠YPX) + (m∠XPZ) = 180 and m∠YPX = 90, then 90 + (m∠XPZ) = 180 |
| 5 | m∠XPZ = 90 | if 90 + (m∠XPZ) = 180, then m∠XPZ = 90 |
| 6 | m∠ZPX = 90 | if m∠XPZ = 90, then m∠ZPX = 90 |
| 7 | ∠ZPX is a right angle | if m∠ZPX = 90, then ∠ZPX is a right angle |
| 8 | m∠ZXY = m∠ZPX | if ∠ZXY is a right angle and ∠ZPX is a right angle, then m∠ZXY = m∠ZPX |
| 9 | m∠YZX = m∠PZX | if m∠YPZ = 180, then m∠YZX = m∠PZX |
| 10 | m∠YZX = m∠XZP | if m∠YZX = m∠PZX, then m∠YZX = m∠XZP |
| 11 | △ZXY ∼ △ZPX | if m∠YZX = m∠XZP and m∠ZXY = m∠ZPX, then △ZXY ∼ △ZPX |
| 12 | △XYZ ∼ △PXZ | if △ZXY ∼ △ZPX, then △XYZ ∼ △PXZ |
| 13 | △PXZ ∼ △XYZ | if △XYZ ∼ △PXZ, then △PXZ ∼ △XYZ |
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