# How can I find the height of an isosceles trapezoid given the upper base length, legs, and a line from the end of a leg to a point on the other leg?

I have an isosceles trapezoid with an upper base length of $$2.6$$ units. I know that the length of the legs is $$3$$ units. I also know that the distance between the bottom of one leg and a point on the opposite leg is $$5$$ units. The point on the opposite leg is $$1$$ unit along the leg from the top ($$2$$ from the bottom). Given these parameters, how can I find the height (or the length of the lower base)?

I recognize that the fact that the trapezoid is isosceles ($$\angle A=\angle B$$), and the fact that $$\overline{BE}=1$$ is necessary to make both the lower base and height definite, but algebraically I cannot understand how to solve for either. I have attempted using law of cosines with two triangles ($$\triangle ABC$$ and $$\triangle BCE$$) and the algebraic relationships between the angles of points on the trapezoid to solve this, but all attempts have led to defining at least one angle as $$0^\circ$$ or $$180^\circ$$, always in a situation when not appropriate.

Using GeoGebra, I found that the lower base is approximately $$5.7$$ units, and the height is approximately $$2.5$$ units. However, I still need to algebraically solve for exact values.

Construction: Drop perpendiculars from points $$A,B,E$$, to base $$CD$$.
Let $$CH=GC=3x \;\; \text{and} \;\; AH=BG=3h$$ Since $$\triangle BGC\sim \triangle EFC$$, $$EF=2h\;,\;FD=2x \;,\; GF=x.$$ Using the Pythagorean Theorem, in $$\triangle EFC$$ and $$\triangle EFD$$, \begin{align*} x^2+h^2&=1\\ (4x+2.6)^2+4h^2&=25 \end{align*} Solving the above equations, we have $$(x,h)=\left(\dfrac{2\sqrt{13\sqrt{109}-95}}{15},\dfrac{2\sqrt{109}-13}{15}\right)$$.
The base length is $$6x+2.6$$ and height is $$3h$$. $$\;\;\;\text{Base}\approx 5.752244$$ $$\text{Height}\approx 2.552613$$