Find $\dfrac {CD-AC} {BE+AC}$ in the convex quadrilateral below

Question

If $AB=ED$ find $\dfrac{CD-AC}{BE+AC}$

(Answer:$1$)

I try

$\triangle ECF \sim\triangle BDF: \dfrac{CD}{BF}=\dfrac{CE}{EF}=\dfrac{DE}{BE} \implies$

$\dfrac{AD-AC}{AD-AF}=\dfrac{BC-CE}{DF-DE}=\dfrac{DF-EF}{AC-CE}$

$\triangle ABC \sim \triangle ADF: \dfrac{AC}{AF}=\dfrac{AB}{AD}=\dfrac{BC}{FD} \implies$

$\dfrac{AD-CD}{AB-BF}=\dfrac{AF+BF}{AC+CD}=\dfrac{BE-CE}{EF+ED}$

Answer 1

Copy $\triangle ABC$, and construct $C'$ so that $\triangle DEC' \cong \triangle ABC$, and $C$ and $C'$ are on different sides of line $DE$.

In quadrilateral $CDC'E$,

• From $\angle DEC' = \angle ABC = \angle ADE$, so $CD \parallel EC'$.
• The opposite vertices $\angle ECD + \angle DC'E = 60^\circ + 120^\circ = 180^\circ$.

So $CDC'E$ is an isosceles trapezium with $60^\circ$ base angles. Its two legs are equal: $CE = C'D$.

By splitting $CDC'E$ into parallelogram $CGC'E$ and equilateral triangle $\triangle GDC'$, all the following lengths are equal:

$$\begin{align*} CE &= C'D = GD\\ &= AC &(\triangle DEC' \cong \triangle ABC) \end{align*}$$

The horizontal sides of parallelogram $CGC'E$ are equal:

$$\begin{align*} CG &= EC'\\ &= BC &(\triangle DEC' \cong \triangle ABC, EC'=BC)\\ CD - GD &= BE + CE\\ CD - AC &= BE + AC\\ \frac{CD-AC}{BE+AC} &= 1 \end{align*}$$

Answer 2

Let point $F$ be intersection of $AB$ and $DE$

$\triangle BEF \sim \triangle DEC$, (AAA Similarity Theorem)

$\frac{BE}{DE}=\frac{EF}{EC}=\frac{BF}{DC}$, (sides are in proportion)

$\frac{BF}{AB}\times \frac{AC}{CD}\times \frac{ED}{FD}=1$ (Menelaus Theorem)

Since $AB=ED$, then $\frac{BF}{AB}\times \frac{AC}{CD}\times \frac{ED}{FD} = \frac{BF \times AC}{CD \times EF}=1$

Since $\frac{EF}{EC}=\frac{BF}{DC}$, and $\frac{BF \times AC}{CD \times EF}=1$, then $AC=EC$

Take the point $H$ on $CD$ such that $AC=CE=CH$

Since $\angle ECH=60^{\circ}$ and $CE=CH$, then $\triangle ECH$ is equilateral.($AC=CH=EC=EH$)

Finally

$BE+AC=BE+EC=BC$ and $CD-AC=CD-CH=DH$

$\angle ACB=\angle EHD=120^{\circ}$

$AC=EH$

$\angle B=\angle D$

$\triangle ABC \cong \triangle EDH$, (SAA Congruency Theorem)

$BC=DH$