Let $ABCD$ be a quadrilateral in a circle with diameter $AC$, and let $DE \perp AB$ with $E$ on $AB$. If $AD = DC$, $[ABCD] = 24$ cm$^2$.

Question

Let $ABCD$ be a quadrilateral in a circle with diameter $AC$, and let $DE \perp AB$ with $E$ on $AB$. If $AD = DC$, $[ABCD] = 24$ cm$^2$. Find $(DE)^2$.

What I Tried: Here is a picture :-

This problem has been eating my brain for quite some time now, and I tried it in many ways but in the end I succeeded doing it. The picture shows it.

First, we have that $BC$ $||$ $ED$ , which is obvious from the information. Next is that since $AD = DC$ , $BD$ is the angle bisector of $\angle BAC$ which gives $BED$ an isosceles triangle and $DE = BE$. Now draw a perpendicular from $C$ to $DE$ and let it intersect at $F$ . We have that $\Delta AED \cong \Delta DFC$ from $RHS$ , and this gives all the necessary piece of info in the picture, also noting that $CFEB$ is a rectangle. This total areas of the $2$ triangles and the rectangle gives the area of the quadrilateral, which looks really good.

From here :-
$$[ABCD] = 2[\Delta AED] + [CFEB]$$
$$\rightarrow 24 = x(x + y) + y(x + y)$$
$$\rightarrow 24 = (x + y)^2$$
$$\rightarrow (x + y) = DE = \sqrt{24}$$

Hence $DE^2 = 24$ .

This solution was so elegant that I couldn't resist posting it in Stack Exchange, maybe others will be helpful with this.

As a note, I will be waiting for some other solutions too 🙂 .

Answer 1

Here is an alternative solution.

First, we note that $ABC$ and $ADC$ are right triangles, and angles $A$ and $C$ in triangle $ADC$ are $45^{\circ}$.

We may also note that $\widehat{DBA}$ faces the same arc as $\widehat{DCA}$ , therefore it should be $45^{\circ}$ too. As a result, in the right triangle $DEB$ we have $DE = EB$ (This interesting information is not part of my solution but I just added it for aesthetic reasons).

Let's add $YXA$ , a rotated copy of the triangle $ABC$ to the shape, as shown in the figure below.

$XA$ is on the same line as $BA$ . We have: $$\widehat{BCA}=\widehat{XAY}$$ $$\widehat{BCA} + \widehat{BAC} = 90^{\circ}$$ It follows that $$\widehat{XAY} + \widehat{BAC} = 90^{\circ}$$ $$\widehat{CAY} = 180 - (\widehat{XAY} + \widehat{BAC}) = 90^{\circ}$$ $$\widehat{DAY} = \widehat{CAY} - \widehat{DAC} = 45^{\circ}$$ The triangles $ADC$ and $ADY$ have an equal angle ($\widehat{DAC} = \widehat{DAY} = 45^{\circ}$) with equal sides ($AC=AY$ and they share $AD$) , therefore they are equal and: $$DY=DC$$ $$\widehat{ADY} = \widehat{ADC} = 90^{\circ}$$ This means that $Y$ , $D$ and $C$ are on the same line. Moreover, this line is divided into equal segments by the three parallel lines $CB$ , $DE$ and $YX$ . In other words, in the trapezoid $CBXY$ , line segment $DE$ , which is parallel to the bases, bisects the legs $CY$ and $BX$ . It follows that $$DE = \frac{1}{2}(CB + YX)$$ And recalling that $YX = AB$ , we have: $$DE = \frac{1}{2}(CB + AB)$$

Now we can move on to the calculation of area. $$S_{ABCD} = S_{ABC} + S_{ADC}$$ $$= \frac{1}{2}(AB.CB + AD^2)$$ $$= \frac{1}{2}(AB.CB + \frac{1}{2}AC^2)$$ $$= \frac{1}{2}(AB.CB + \frac{1}{2}(AB^2 + CB^2))$$ $$= \frac{1}{4}(AB + CB)^2$$ $$= DE^2$$ Voila: $S_{ABCD} = DE^2$ ! (This is an exclamation mark, not a factorial)