Area from a Perpendicular Sum

As ABCD is a cyclic quadrilateral, by the property that opposite angles are supplementary, ∠BCD = 90° the perpendicular to either of DC and CB will be in the exterior of the quadrilateral. We can see it by observation. Now after drawing the perpendiculars to AE and AF, we get a quadrilateral AFCE in which ∠AEC =∠ECF = ∠AFC = 90° . So by angle sum property of a quadrilateral, ∠FAE = 90° . So this quadrilateral turns out to be a parallelogram. [since the angles are right, it is a rectangle - Calvin]

We can also see that ΔAEB and ΔAFD are formed. We can show that these two Δs are similar by the Angle-Side-Angle congruency criterion. This implies that the perpendiculars AE and AF are also congruent. It means each of them is 8 and the quadrilateral AFCE is a square with area = 64. Now [ABCD]=[AFCE]-[AFD]+[AEB] . But since AEF and AEB are congruent, we get [ABCD]=[AFCE] Therefore, [ABCD]= 64

We know that ABE and AFD are similar because <EAF = <BAD = 90 degrees. AB = AD, so AEB is congruent to AFD (and AE = AF). Thus the area of ABCD is the area of AECF. Then [AECF] = AE AF = (16/2) (16/2) = 64.

Let's take a closer look at $AECF$ . We have $\angle ECF = \angle BCD = 180^\circ - \angle BAD = 90^\circ$ . Hence, $AECF$ is a rectangle and $\angle EAF = 90^\circ$ , which gives $\angle BAE = \angle DAF$ . Thus $AE = AB \cos \angle BAE = AD \cos \angle FAD = AF$ , hence it is a square.

Furthermore, $ABE$ is congruent to $AFD$ , so $[ABCD] = [AECF] = 8 ^2 = 64$ .

For this problem we know that $AB=AD$ and that $\measuredangle BAD=90°$ .

We also know that the feet E and A lie on the lines BC and DC.

Since this is the case we can deduce the following.

$\measuredangle ABC\quad \le \quad 90°$

$\measuredangle ADC\quad \le \quad 90°$

Since this is a cyclical quadrilateral, the angles ABC and ADC cannot both be less than 90 degrees therefore they must equal 90 degrees and the cyclical quadrilateral we're dealing with is simply a square.

This means that foot E lies at point B and foot F lies at D.

Therefore

$AB=AD=AE=AF$

Since $AB=AD$ and $AE+AF=16$ , $AB$ and $AD$ are equal to 8.

Thus the area is simply $A=s^2$ which is 64.

Ptolemy's theorem expresses the product of the lengths of the two diagonals p and q of a cyclic quadrilateral as equal to the sum of the products of opposite sides:

pq = ac + bd.

also here both diaginals are equal to the diameter as it has angle A=90 and angle inscribed in a semicircle is 90

so by ptolemy thm ac+bd=ad+bc ..1

now let AB(a)=AD(c)=x then BD=p^2=2x then from 1 a=d similarly we can say a=b=c=d=x also one angle 90 hence a square sum of 2 sides=16 each side=8 hence area =64

Since ABCD is a cyclic quadrilateral, ∠BAD + ∠BCD = 180∘ And, since ∠BAD is 90∘, ∠BCD is also 90∘. Therefore, to make ABCD a cyclic quadrilateral with 2 opposite angles, equalling to 90∘ each, the other 2 angles also have to be 90∘ and has to be a square. Since E and F are foots of perpendicular from A to BC and A to DC respectively, E is actually B, F is actually D. Hence, AE + AF = AC + AD = 16. As ABCD is a square, AB = AC = 8. To find the area of square, it would be length x length = 8 x 8 = 64

Given \angle BAD = 90 ^ \circ In a cyclic quadrilateral opp. angles are complementary. Thus, \angle BCD=90^ \circ. Given AB=AD. In the quadrilateral, two adjacent sides are equal and opp. angles are 90^ \circ. Thus, it is a square. Then, AE+AF=16 gives sum of its sides. Therefore, side of the square=8. Area=64