check the picture.
In triangle ABC:
m(B)=90°, let m(A)=α, and let m(C)=β
so m(B)+m(A)+m(C)=180°, so α+β=90°
In triangle ABD, m(A)=α, m(ADB)=90° so m(ABD) must be β, so that the sum completes to 180.
By the same logic, m(DBC)=α
so we have the following similarities of triangles: (let's go by the order α-β-90°):
ACB ≡ ABD ≡ BCD,
consider ABD ≡ BCD:
[tex] \frac{AB}{BC} = \frac{BD}{CD} = \frac{AD}{BD} [/tex]
substitute with the givens:
[tex] \frac{e}{h} = \frac{j}{40} = \frac{27}{j} [/tex]
from
[tex]\frac{j}{40} = \frac{27}{j}[/tex]
we have
[tex] j^{2}=40*27 =1080[/tex]
In triangle ABD, from the pythagorean theorem:
[tex] e^{2} = j^{2}+27^{2}=1080+729=1089[/tex]
so [tex]e= \sqrt{1089}= 42.53 (units)[/tex]
Remark: a direct solution can be given by Euclid's theorem, AB^2=AD*AC
[tex]AB^2=AD*AC[/tex] so
[tex] e^{2}=27*47[/tex], then take square root of e
Answer: 42.53 units
