Answer:
By exterior angle theorem, we have;
∠DBE = ∠H + ∠HEB = ∠ECD = ∠H + ∠HDC
∴ ∠H + ∠HEB = ∠H + ∠HDC
By addition property of equality, we have
∠HEB = ∠HDC
∠H = ∠H by reflexive property
∴ ΔHCD ~ ΔHEB by Angle Angle AA similarity postulate
∴ HE/HD = EB/DC, by the definition of similarity
Therefore, by cross multiplication, we have;
HE × DC = EB × HD
Therefore, by commutative property of multiplication, we have;
HE × DC = HD × EB
Step-by-step explanation:
When dilation is about the origin, as it is here in every case, the image point coordinates are the original (pre-image) coordinates multiplied by the scale factor.
1. Multiply every coordinate value by 5:
... W' = (-5, 10), X' = (-15, -5), Y' = (25, -5), Z' = (15, 10)
2. Multiply every coordinate value by 1/3:
... A' = (-2, 5), B' = (0, 5/3), C' = (1, 10/3)
3. A' = (2, 8), B' = (6, 2), C' = (2, 2)
4. The image coordinates are 5 times the original coordinates, so ...
... the scale factor of the dilation is 5.
<span>(4,5); r= square root of 12</span>
