Answer: The length of segments between this point and the vertices of greater base are [tex]7\frac{5}{7}[/tex] and 18.
Step-by-step explanation:
Let ABCD is the trapezoid, ( shown in below diagram)
In which AB is the greater base and AB = 18 DC= 11, AD= 3 and BC = 7
Let P is the point where The extended legs meet,
So, according to the question, we have to find out : AP and BP
In Δ APB and Δ DPC,
∠ DPC ≅ ∠APB ( reflexive)
∠ PDC ≅ ∠ PAB ( By alternative interior angle theorem)
And, ∠ PCD ≅ ∠ PBA ( By alternative interior angle theorem)
Therefore, By AAA similarity postulate,
[tex]\triangle APB\sim \triangle D PC[/tex]
Let, DP =x
⇒ [tex]\frac{3+x}{18} = \frac{x}{11}[/tex]
⇒ 33 +11x = 18x
⇒ x = 33/7= [tex]4\frac{5}{7}[/tex]
Thus, PD= [tex]4\frac{5}{7}[/tex]
But, AP= PD + DA
AP= [tex]4\frac{5}{7}+3 =7\frac{5}{7}[/tex]
Now, let PC =y,
⇒ [tex]\frac{7+y}{18} = \frac{y}{11}[/tex]
⇒ 77 + 11y = 18y
⇒ y = 77/7 = 11
Thus, PC= 11
But, PB= PC + CB
PB= 11+7 = 18
