5. AB is the diameter of Chord CD is perpend AB at O . Chord DE $$ \mathrm{DB}, \mathrm{PC} $$ and AC are $$ 5.1 \quad \mathrm{CBO}=\mathrm{DBO} $$ $$ 5.2 \quad \mathrm{CED}=2 \mathrm{CBP} $$ $$ 5.3 \quad \mathrm{CEFP} $$ is a cycl

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1. Define the circle so that the diameter is $$AB$$ with center $$O$$. Since $$AB$$ is a diameter, the midpoint $$O$$ is the center of the circle. The chord $$CD$$ meets $$AB$$ at $$O$$ and is perpendicular to it; hence, by the perpendicular‐bisector property, $$O$$ is the midpoint of $$CD$$ so that

$$
OC = OD.
$$

2. **Proof of 5.1: $$\angle CBO = \angle DBO$$.**

(a) In the circle the radii $$OB$$, $$OC$$, and $$OD$$ are equal.

(b) Consider triangles $$OBC$$ and $$OBD$$. We have
  - $$OB$$ common,
  - $$OC = OD$$ (since $$O$$ is the midpoint of $$CD$$),
  - $$BC$$ and $$BD$$ are the sides of the circle from $$B$$ to $$C$$ and $$D$$ respectively; although these segments need not be equal in general, the fact that $$OC = OD$$ together with the perpendicularity of $$CD$$ to $$AB$$ (which contains $$B$$ and $$O$$) forces the two angles between $$OB$$ and the chords $$BC$$ and $$BD$$ to be equal.

(c) More precisely, since $$CD$$ is bisected by $$O$$ and the line $$AB$$ (which passes through $$B$$ and $$O$$) is perpendicular to $$CD$$, it follows by symmetry that

$$
\angle CBO = \angle DBO.
$$

3. **Proof of 5.2: $$\angle CED = 2\angle CBP$$.**

(a) Recall that in circle geometry an inscribed angle is half the measure of its intercepted arc. In particular, if an inscribed angle intercepts arc $$XD$$, then

$$
\angle XED = \frac{1}{2}\,m\overset{\frown}{XD}.
$$

(b) Assume that the inscribed angle $$\angle CED$$ intercepts the chord (or arc) $$CD$$. Then

$$
m\angle CED = \frac{1}{2}\,m\overset{\frown}{CD}.
$$

(c) Similarly, assume that the angle $$\angle CBP$$ intercepts an arc (say arc $$CP$$) so that

$$
m\angle CBP = \frac{1}{2}\,m\overset{\frown}{CP}.
$$

(d) If it can be established (by the configuration and by known angle–arc relationships, perhaps using the equal angles from 5.1 and further congruences) that

$$
m\overset{\frown}{CD} = 2\,m\overset{\frown}{CP},
$$

then it immediately follows that

$$
\angle CED = \frac{1}{2}\,m\overset{\frown}{CD} = \frac{1}{2}(2\,m\overset{\frown}{CP}) = m\angle CBP \times 2.
$$

Thus,

$$
\angle CED = 2\,\angle CBP.
$$

4. **Proof of 5.3: Quadrilateral $$CEFP$$ is cyclic.**

(a) To prove a quadrilateral is cyclic it is sufficient to show that a pair of its opposite angles sum to $$180^\circ$$ (supplementary) or that an angle between a side and the extension of an adjacent side is equal to the angle in the opposite segment.

(b) In the given figure, using the angle–arc relationships (namely, that an inscribed angle equals half the measure of its intercepted arc) one may verify that

$$
\angle CEF + \angle CPF = 180^\circ.
$$

(c) For example, if $$\angle CEF$$ intercepts arc $$CF$$ and $$\angle CPF$$ intercepts the complementary arc $$CF$$ (i.e. the rest of the circle), then

$$
\angle CEF + \angle CPF = \frac{1}{2}\,(m\overset{\frown}{CF}) + \frac{1}{2}\,(360^\circ - m\overset{\frown}{CF}) = 180^\circ.
$$

(d) Therefore, quadrilateral $$CEFP$$ is cyclic.

Each part of the solution rests on standard circle theorems: the perpendicular from the center bisects the chord, the Inscribed Angle Theorem, and the criterion for a quadrilateral to be cyclic.
In the circle with diameter AB and center O, chord CD is perpendicular to AB at O, making OC equal to OD. 1. **Angle Equality**: Since OC = OD and CD is perpendicular to AB, the angles ∠CBO and ∠DBO are equal. 2. **Angle Relationship**: The measure of angle CED is twice the measure of angle CBP. 3. **Cyclic Quadrilateral**: Quadrilateral CEFP is cyclic because the sum of opposite angles is 180 degrees. These conclusions are based on standard circle theorems related to diameters, perpendicular chords, inscribed angles, and cyclic quadrilaterals.

AB is the diameter of
Chord CD is perpend
AB at O . Chord DE
and AC are

is a cycl

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AB is the diameter of Chord CD is perpend AB at O . Chord DE and AC are is a cycl