As shown in the figure, five pieces equilateral triangle of the same size are put in order without a gap.
Among quarters point of BC, point D is the point closest to the B.
The length of AE is 9 cm.
Find the length of AD.
Among quarters point of BC, point D is the point closest to the B.
The length of AE is 9 cm.
Find the length of AD.
Answer
In Fig. 1, since AH and BC are the diagonal lines of rhombus ABHC, ∠AGD = ∠AGF = 90°.
9/4 cm
Solution
In Fig. 1, △ACF and △BEF are homothetic and a homothetic ratio is AC : BE = 1 : 3.
CF : FB = 1 : 3 as well.
The length of AF is 9cm × 1 / (1+3) = 9/4 cm.
D is a point closest to B among quarters point of BC.
Furthermore, since it is BG = CG, according to Fig.2, it is found that BD = DG = GF = FC.
In Fig. 1, since AH and BC are the diagonal lines of rhombus ABHC, ∠AGD = ∠AGF = 90°.
Since as for △ADG and △AFG, these are right triangles with DG = GF and AG = AG, △ADG and △AFG are congruent.
Thus, AD = AF.
Therefore, AD = AF = 9/4 cm.
