The triangles ΔCOP and ΔCO''P'' can be shown to be similar when OP is
equal to O''P''.
- The true statement is option D. ΔCO''P'' is a dilation of ΔCOP with center C and a scale factor, r = [tex]\underline{\dfrac{C'O'}{CO} = \dfrac{CO''}{CO}}}[/tex]
Reasons:
The given parameters are;
∠C ≅ ∠C'
∠POC ≅ ∠P'O'C'
The point O'' is on [tex]\overline{CO}[/tex]
CO'' = C'O'
The point P'' is on [tex]\overline{CP}[/tex]
Required:
The true statement from among the given statement
Solution:
Statement [tex]{}[/tex] Reason
∠C ≅ ∠C' [tex]{}[/tex] Given
∠COP ≅ ∠C'O'P' [tex]{}[/tex] Given
ΔCOP ~ ΔC'O'P' [tex]{}[/tex] AA similarity postulate
ΔC'O'P' is a dilation ΔCOP [tex]{}[/tex] Similar but not congruent triangles
CO'' = C'O' [tex]{}[/tex] Given
CO'' ≅ C'O' [tex]{}[/tex] Definition of congruency
P'' is a dilation of P [tex]{}[/tex] Given
Therefore, where CP'' = C'P', we have;
ΔCO''P'' ≅ ΔC'O'P' [tex]{}[/tex] SAS rule of congruency
ΔCO''P'' = ΔC'O'P' [tex]{}[/tex] Definition of congruency
ΔCOP ~ ΔCO''P'' [tex]{}[/tex] Transitive property
ΔCO''P'' is a dilation ΔCOP [tex]{}[/tex] Similar but not congruent triangles
Which gives;
[tex]\mathrm{\Delta CO''P'' \ is \ a \ dilation \ of \ \Delta COP \ with \ center \ C \ and \ scale \ factor \ r = \dfrac{C'O'}{CO} = \dfrac{CO''}{CO}}[/tex]
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