First, we assume that we can draw a perpendicular line from X to YZ.

Notice that PYX and XYZ share the angle PYX. We use this fact to show that △PYX and △XYZ are similar triangles.

Similarly, PXZ and XYZ share the angle PZX. Thus, we can show that △PXZ and △XYZ are also similar triangles.

Then:

$\frac{\mathrm{XY}}{\mathrm{ZY}}=\frac{\mathrm{YP}}{\mathrm{YX}}$

Where XY stands for "distance between X and Y."

We also claim that:

$\frac{\mathrm{XZ}}{\mathrm{YZ}}=\frac{\mathrm{PZ}}{\mathrm{XZ}}$

Then using the cross multiplication theorem, we make the following claim:

XY ⋅ XY = YP ⋅ YZ

XZ ⋅ XZ = PZ ⋅ YZ

We can add the left sides and the right sides to conclude that:

XY ⋅ XY + XZ ⋅ XZ = YP ⋅ YZ + PZ ⋅ YZ

Using the distributive property, we claim that:

YP ⋅ YZ + PZ ⋅ YZ = (YP + PZ) ⋅ YZ

Since YP + PZ = YZ:

YP ⋅ YZ + PZ ⋅ YZ = (YZ) ⋅ YZ

Finally, using the transitive property, we claim that:

XY ⋅ XY + XZ ⋅ XZ = YZ ⋅ YZ

Quiz (1 point)

• (a ⋅ b) + (a ⋅ c) = a ⋅ (b + c)
• if (∠ZXY is a right angle) and (∠XPY is a right angle) and (m∠YPZ = 180), then △PYX ∼ △XYZ
• if (∠ZXY is a right angle) and (∠XPY is a right angle) and (m∠YPZ = 180), then △PXZ ∼ △XYZ
• if △ABC ∼ △XYZ, then (distance CB) / (distance ZY) = (distance BA) / (distance YX)
• if △ABC ∼ △DEF, then (distance BC) / (distance EF) = (distance CA) / (distance FD)
• if ((distance AB) / (distance CB) = (distance BX) / (distance BA)) and ((distance AC) / (distance BC) = (distance CX) / (distance CA)), then ((distance AB) ⋅ (distance AB)) + ((distance AC) ⋅ (distance AC)) = ((distance BC) ⋅ (distance BX)) + ((distance BC) ⋅ (distance CX))
• if m∠ABC = 180, then (distance AB) + (distance CB) = distance AC

• if the following are true:

  • a = b ⋅ c
  • c = d

  then a = b ⋅ d

• if the following are true:

  • a = b
  • b = c

  then a = c

Please write your proof in the table below. Each row should contain one claim. The last claim is the statement that you are trying to prove.

Step	Claim	Reason (optional)	Error Message (if any)
1
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