Day 3: Just Enough Information

Firstly $\angle ADC = \angle ABC = 90^{\circ}$ . Let $AD = a, DC = b$ for clarity. Then the area of $ADC = \frac12 ab$ .

As $AB$ and $BC$ subtend the same angle they are of equal length. So the area of $ABC = \frac14 AC^2$ , since if you construct the square with side $AC$ , then it is clear that $ABC$ has one quarter of that area.

So the area of $ABCD = \frac14 AC^2 +\frac12 ab$ .

But $AC^2 = a^2 + b^2$ , so: $\mathrm{Area\ of\ ABCD} = \frac14 (a^2 + b^2 +2ab) = \frac14 (a+b)^2 = \boxed{625}$

It is amazing that the area is independent of $a$ and $b$ ! Also for this to be a complete solution we must show that such a quadrilateral is possible; a square with sidelength $25$ works perfectly.

Angles at D add up to 90^o, it is clear that AC is the DIameter. So ABC=90^o .
Angles by chord BC,, CDB=CAB=45^o. So BCA=45^O.
Angles ADB=BDC, so chords AB=BC..
Angles by chord AD, CDB=CAB=45^o. Will soon complete the solution.

Its fine to assume that point D lies anywhere on the arc AC, so here's a numerical approach: Let AD = 10 and CD = 40. Since triangle ACD is a right triangle, AC^{2} = 100 + 1600 = 1700 Since triangle ABC is another right angled isoceles triangle, then AB = BC = sqrt{(1700/2}= sqrt{850}

Hence total area = area of both triangles = 1/2 AD×BC×sin90 + 1/2 AD × CD×sin90 = 1/2×850×1 + 1/2×10×40×1 = {625}