Can this be solved without trigonometry?

Question

My Attempt: The triangle is equilateral so all sides are equal let them be $= l $ $$\angle CAD = 90^\circ-60^\circ =30^\circ$$ $$ \angle ACD = 60^\circ$$ After this don't know how to find $CD$ but I know what to do if I manage to find $CD$

Answer 1

In case you want the proof absolutely without Trigonometry, I found a way which uses area and Pythagorean theorem.

Cleary $Ar(\Delta ABC) + Ar(\Delta ADC) = Ar(Trap. ABCD) $

So, $\frac{\sqrt3}{4}AB^2+\frac{1}{2} AB·CD=\frac{1}{2}·AD(CD+AB)$

Simplifying we have $$\frac{\sqrt3}{2}AB^2+CD·AD=AD·CD+AB·AD$$ $$AD=\frac{\sqrt3}{2}AB$$ But $AC=AB$

So, $$CD^2+AD^2=AC^2=AB^2$$ Put $AD=\frac{\sqrt3}{2}AB$

So, $$CD^2+\frac{3}{4}AB^2=AB^2$$ This yields $$CD=\frac{AB}{2}$$

Now, $ CD + AB = 2·EF = 1.5 a$

Putting $CD=\frac{AB}{2}$

$ \frac{3}{2}AB = 1.5 a$

$AB = a$

Answer 2

I found quite a nice ''visual'' way of figuring out AB. Hope this clears up the previous proof. I only used Pythagoras theorem.

Answer 3

Proof without words shows that $AB=4\frac{a}{4}=a.$

Answer 4

$DC=\frac12 AB$ and $EF$ is the mean of $AB$ and $DC$, which is $\frac34 AB$

So $$\frac34 AB=0.75a\implies AB=a$$

Answer 5

Welcome to Maths Stack Exchange!

Here's a very simple way to solve the problem without guessing.

Firstly, construct $CH\perp AB$, with $H$ on $AB$. As $\triangle ABC$ is equilateral, $AH=BH=\frac12 AB$.

We also have $\angle CHA=90^\circ$, and from the problem, $\angle CDA=\angle DAB=90^\circ$, and so all four angles in quadrilateral $CDAH$ are $90^\circ$, as the angle sum of a quadrilateral is $360^\circ$.

This means that $DC=AH=\frac12AB$. We can also use the fact that the average of the lengths of the two bases of a trapezium is equal to the length of its midline. Since $EF$ is the midline of trapezium $ABCD$, we have

$$EF=\frac{CD+AB}{2}$$

Substituting the values in, we get

$$0.75a=\frac{\frac12 AB+AB}{2}$$ $$1.5a=1.5AB$$ $$AB=a$$