Four Points on a Semicircle

$AB$ subtends a right angle at $C$ (or $D$ ). By pythagoras, $AC^2 = AB^2 - BC^2 = 160000 - 10000 = 150000$ , $AC = 100 \sqrt {15} = BD$ . Let $DC = x$ . Now by applying Ptolemy's Theorem to the cyclic quadrilateral $ABCD$ , We have $AD*BC + AB*CD = AC*BD$ so $(100)(100) + (400)(x) = 150000$ . $x = 350$ on simplification.

[LaTeX edits - Calvin]

Let the centre of the circle be point $E$ . By the Law of Cosines, the angle $\angle AED = \arccos\frac{200^2+200^2-100^2}{2\left(200\right)\left(200\right)} = \arccos\frac{7}{8}$ . Thus, $\angle DEC = 180-2\arccos\frac{7}{8}$ . Applying the Law of Cosines one more time yields $DC = \sqrt{200^2+200^2-2\left(200\right)\left(200\right)\cos\left(180-2\arccos\frac{7}{8}\right)} = 350$ .

In this question, we have a cyclic quadrilateral. According to Ptolemy's theorem, sum of, product of, lengths of opposite sides is equal to product of lengths of diagonals of the cyclic quadrilateral. Here, we have, AC \times BD = AB \times DC + AD \times BC .......(1) When we draw the diagonals we get two right angled triangles, namely triangle ADB (right angled at D since, angle in a semi-circle is a right angle). and triangle ACB right angled at C. By Pythagoras theorem we have value of diagonals as \sqrt{400^2 - 100^2} which will be \sqrt{150000} . Hence product of diagonals = 150000. On substituting in (1) we have 150000 = 400 DC +100 100 i.e., 140000 = 400*DC which gives us the value of DC as 350.

Let $AC$ and $BD$ intersect at $X$ . Then let $DX = CX = a$ , and $AX = BX = b$ . Due to the properties of a semicircle, it is clear $\angle ACB = \angle ADB = 90^\circ$ , and as such, we may find via Pythagoras' Theorem that: (1) $a^2 + 100^2 = b^2$ (2) $(a+b)^2 + 100^2 = 400^2$

Solving these two simultanously, we are able to find that $\frac{a}{b} = \frac{7}{8}$ . Finally, we note that by similar triangles $\triangle ABX$ and $\triangle CXD$ , we have that $\frac{a}{CD} = \frac{b}{400}$ , and hence $CD = 400 (\frac{a}{b})$ . And hence our answer is $350$ .

Ptolemy's Theorem tells us that AB * DC + AD * BC = AC * BD

Since C and D are points on a semicircle with diameter AB , triangles ACB and ADB are right; they're also congruent, with AC = BD = sqrt{400^2 - 100^2} = sqrt{150000} by Pythagoras.

Then AB * DC + AD * BC = sqrt{150000}^2, and by plugging in known values, we get that 400 * DC = 150000 - 10000 = 140000, or that DC = 350.

Let $O$ be the centre of the semicircle. Now join $OD$ & $OC$ . Now draw a perpendicular from $ED$ on $AO$ and $CF$ on $OB$ . [He likely means "drop a perpendicular from $D$ to $AO$ intersecting at $E$ ". - Calvin] Now in Right angled triangle $OED$ , we have $(OE)^2+(ED)^2=(OD)^2$ ( $OD$ =radius of the circle=200). Similarly in Right angled triangle $AED$ , $(AE)^2+(ED)^2=(AD)^2$ (given AD=100).

Thus $(200-OE)^2+(ED)^2=(100)^2 (ED)^2=(100)^2-(200-OE)^2$ . Now putting the value of $(ED)^2$ in the equation of triangle $OED$ , we get, $(100)^2-(200-OE)^2+(OE)^2=(400)^2$ . By solving this we get $OE=175$ . Now in triangle in $AOD$ & $BOC$ , since $OA=BO$ , $\angle AOD=\angle BOC$ and $OD=OC$ , so triangle $AOD$ & $BOC$ are congruent by SAS(side-angle-side) congruency. so $OE=OF$ and hence $2(OE)=EF=350$ . Now $EF=DC=350$ .

[Latex Edits. Edits for clarity - Calvin]

If $O$ is the midpoint of $AB$ , let $\alpha$ be the measure of $\angle AOD$ .

By bisecting $\angle AOD$ , observe that

• $\sin \frac{\alpha}{2} = \frac{\overline{AD}}{2 \overline{AO}} = \frac{1}{4}$
• $\cos \frac{\alpha}{2} = \sqrt{1-(\frac{1}{4})^2}=\frac{\sqrt{15}}{4}$ .

and so, $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}= \frac{\sqrt{15}}{8}$ .

Now let $P$ be the midpoint of $DC$ . Clearly $DC$ is parallel to $AB$ and $\angle AOP$ is a right angle, so $\angle AOD$ and $\angle DOP$ are complimentary. Therefore, if $\beta$ is the measure of $\angle DOP$ , we have $\sin \beta = \cos \alpha = \sqrt{1- (\frac{\sqrt{15}}{8})²} = \frac{7}{8}$ .

Finally, $\overline{DC}=2 \overline{DP}=2 \overline{OD} \cdot \sin \beta=350$ .

as ABCD cyclic let DC=x , 100 100+400 x=400^2-100^2, it follows x=350 immidiately.

using properties of cyclic quadrilateral ac+bd=pq where a,b,c,d are side lengths and p,q are length of the two diagonals. Lengths of the two diagonals can be found using pythagoras theorem(angle in a semicircle is right angle).

DC = AC cos(theta1) - AD cos(theta2) = AC (AC/AB) - AD (AD/AB) = (AC^2 - AD^2)/AB = (AB^2 - 2 AD^2)/AB = (400^2 - 2 100^2)/400 = Ans

Solution 1: Since $AB$ is the diameter, we have $\angle ADB = 90^\circ$ . Drop a perpendicular from $D$ to $AB$ at $E$ . So, we have triangles $AED$ and $ADB$ are similar by angle-angle-angle. Thus $\frac{AE}{AD} = \frac{AD}{AB}$ $\Rightarrow AE = \frac{100\cdot 100}{400} = 25$ .

Let $O$ be the center of the semicircle. By symmetry, we have $DC = 2\cdot OE = 2 \cdot (OA - EA) = 2\cdot (200 - 25) = 350$ .

Solution 2: Since $AD = BC$ , so $\angle DBA = \angle CAB = \angle CDB$ , so $AB \parallel DC$ , which makes $ABCD$ an isosceles trapezium.

Let $O$ be the center of the semicircle. Then $OAD$ and $OBC$ are congruent isosceles triangles. Let $\angle DAO = \alpha$ . Then, $\angle ADO = \alpha$ , $\angle DOA = 180^\circ - 2 \alpha$ , $\angle ODC = 180^\circ - 2\alpha$ , $\angle COB = 180^\circ - 2\alpha$ , $\angle DOC = 4\alpha - 180^\circ$ .

By sine rule, $\frac {AD}{\sin (180^\circ-2\alpha) } = \frac {AO} { \sin \alpha} \Rightarrow 4 \sin 2 \alpha = \sin \alpha \Rightarrow 4 \sin \alpha \cos \alpha = \sin \alpha$ . Since $\alpha \neq 0$ , so $\sin \alpha \neq 0$ and thus $\cos \alpha = \frac {1}{4}$ .

By sine rule, we have $\frac {DC}{\sin (4\alpha - 180^\circ)} = \frac {OC}{\sin(180^\circ - 2 \alpha)}$ $\Rightarrow DC = \frac {200 \times (- \sin 4\alpha)} {\sin 2 \alpha}$ $=- 200 \times 2 \times \cos 2 \alpha$ $= -400 \left( 2 \left(\frac {1}{4}\right)^2 - 1 \right)$ $= -400 \left( - \frac {7}{8} \right)$ $= 350$ .

Note: There is another solution by Pythagorean Theorem + Ptolemy's Theorem.

We can solve it by using geometric mean Assume that AB=CD+2x

Because ACB is a right triange .. (BC)^2=x.(AB) 100^2=400x x=25

So

CD=AB-2x=400-2(25)=350