Can't you draw the other altitude?

Since $\angle AHD = \angle CHE$ , as they are opposite angles, and $\angle ADC = \angle AEC = 90^\circ$ , $\angle BAE = \angle BCD$ . And since $\angle AEB = 90^\circ$ , $\angle ABC = \angle CHE$ and $\triangle ABE$ is similar to $\triangle CHE$ . Therefore, $\dfrac {HE}{EC} = \dfrac {BE}{AE}$ $\implies \dfrac {HE}2 = \dfrac 34$ $\implies HE = \dfrac 34 \times 2 = \boxed{1.5}$ .

Draw the other altitude named $BF.$ Let $HE=x.HC=\sqrt{x^2+4},HB=\sqrt{x^2+9},AB=5$ and $AC=2\sqrt5.$ [By pythogoras theorem]. $AF=FC=\sqrt5$ since $\triangle ABC$ is isosceles and the altitude of an isosceles triangle is a median also.
$FH=\sqrt{x^2-1}$ [By pythogoras theorem]. $AH=\sqrt{x^2+4}$ [By pythogoras theorem].
$AH+HE=AE.$ So $\sqrt{x^2+4}+x=4\Rightarrow {(\sqrt{x^2+4})}^2={(4-x)}^2\Rightarrow x=\boxed{1.5}.$