Subject: Geometry

# Circles

• If a line is tangent to a circle, then it is perpendicular to the radius from the point of tangency. line l is perpendicular to the radius \overline{AB}
• Two tangents from a common external point are congruent. \overline{WY} and \overline{WZ} are tangent to ⊙X meeting at a common external point at point W. Based on the corrollary that two tangents from a common external point are congruent, it can be concluded that \overline{WY}\cong \overline{WZ} . Suppose the length of \overline{WZ} is equal to 5, what could be the length of \overline{WY} ? The length of \overline{WY} is also equal to 5 since \overline{WY}\cong \overline{WZ} .
• A line in the plane of a circle which is perpendicular to a radius at its endpoint on the circle, is tangent to the circle. In the image above, \overleftrightarrow{l} is perpendicular to \overline{AB} . Therefore, \overleftrightarrow{l} is tangent to \bigodot{A} .
• The measure of an arc formed by two adjacent non-overlapping arcs, is equal to the sum of the measure of the two arcs. On the image above, arc BD is formed by arc BC and arc CD. Based on the postulate, the measure of arc BD is the sum of the measures of arc BC and arc CD, that leads us to this formula: BC+CD=BD. Given on the image above, m BC is 45o; m CD 55o, by substitution we will then have 45o+55o= 100o. Thus, the measure of arc BD=100o.
• In the same circles or in congruent circles, if two chords are congruent, then their intercepted arcs are also congruent. In the image above, if you notice, 2 triangles are formed namely \Delta{PSQ} and \Delta{RSQ} with a common side \overline{QS} . Given, \overline{PQ}\cong\overline{QR} ; Radiis in a circle are congruent, so we can say that \overline{PS} is also congruent with \overline{RS} . Since the two triangles have a common side \overline{QS} , \Delta{PSQ}\cong\Delta{RSQ} by the SSS postulate (all three sides of \Delta{PSQ}\cong\Delta{RSQ} ). Since \Delta{PSQ}\cong\Delta{RSQ} , we can also say that \angle{PSQ} is equal to \angle{RSQ} . The measure of an arc depends on the measurement of its central angle, thus \angle{PSQ} =m arc PQ , \angle{RSQ} = m arc QR. \angle{PSQ} is equal to \angle{RSQ} , thus arc PQ=arc QR. As the segments PQ and QR are equal, it follows that arc PQ is also equal to arc QR.
• In the same circle or in congruent circles, if two arcs are congruent, then their chords are also congruent. In the image above, arc BD \cong arc CD, it follows then that \overline{BD}\cong\overline{CD}
• The diameter of a circle that is perpendicular to a chord, bisects the chord and its arc. The diameter \overline{SP} is perpendicular to a chord \overline{RT} . Based on the theorem, at is it perpendicular to \overline{RT} , it also bisects it as well as its arc, giving us: \overline{QR}\cong\overline{QT} and arc PR is equal to arc PT.
• If two chords are congruent, then they are equidistant from the center./ if two chords are equidistant from the center, then they are congruent. \overline{BC}\cong\overline{DE} . Based on the theorem, they are also equidistant. Thus, \overline{AF}\cong\overline{AG} .
• The measure of an inscribed angle is one-half the measure of its intercepted arc. In the image above, \angle{CAD} intercepts arc CD. To find the measure of \angle{CAD} , based on the theorem, we will then have m\angle{CAD} =1/2 of the measure of arc CD. Given the measure of arc CD with 50o, the m\angle{CAD} =1/2(50o). Thus, m\angle{CAD} =25o.
• If two inscribed angles intercept the same arc or congruent arcs, then the angles are congruent In the image above \angle{a} and \angle{b} both intercept arc CD, thus, \angle{a}\cong\angle{b}
• If two arcs of a circle are included between parallel chords, then the arcs are congruent Arc PQ and Arc RS are positioned between \overline{PQ} and \overline{RS} , which are parallel to each other, thus arc PQ is \congarc RS.
• If an inscribed angle intercepts a semi-circle, then the angle is a right angle \angle{W} intercepts arc YZ which is a semi-circle, thus \angle{W} is a right angle.
• If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary MOPN is a quadrilateral inscribed in \bigodot{L} , wherein \angle{M} is opposite to \angle{P} and \angle{O} is opposite to \angle{N} . Thus, \angle{L} and \angle{P} are supplementary, and so with \angle{O} and \angle{N}
• The measure of the angle formed by a chord and a tangent, is equal to one-half the measurement of the intercepted arc. \angle{ACD} is formed by the chord \overline{AC} and tangent \overrightarrow{CD} . Given these parts, \overline{BC}\perp\overline{CD} since tangents are always perpendicular to the radius from the point of tangency. This makes [$\angle{ACD}$} = 90o. \overline{AC} which is the diameter , forms arc AC, which is equal to 180o. It follows then that, the m\angle{ACD} =1/2 to the measure of arc AC.
• The measure of each angle formed by two chords that intersect at an interior point of the circle,is equal to one half the sum of the measures of the intercepted arcs. \bigodot{P} has chords \overline{TU} and \overline{RS} which intersects at an interior point Q. This forms \angle{TQS} , \angle{RQU} , \angle{TQR} and \angle{SQU} . Based on the theorem, to be able to get the measures of these angles, we have to multiply sum of the measures of the arc it intercepts. Suppose we want to get m \angle{TQS} . It intercepts arc TS and arc RU, thus: m \angle{TQS} =1/2 (measure of arc TS+ measure of arc RU).
• The measure of the angle formed when two secants, two tangents or a tangent and a secant intersect in a point in the exterior of a circle,is equal to one-half the difference of the measures of the intercepted arcs. \bigodot{D} has secant \overline{AC} and tangent \overrightarrow{AB} . It forms \angle{BAC} .

Based on the theorem the m \angle{BAC} = 1/2 (measure of arc BC - measure of arc BE).

• When two chords intersect within a circle, then the product of the lengths of one chord is equal to the product of the lengths of the other chord. \bigodot{A} has chords \overline{CF} and \overline{DE} which intersects at an interior point B.

Based on the theorem, it follows that \overline{BC} x \overline{BF} = \overline{BE} x \overline{BD}

• When two secant lines of the same circle intersect at an exterior point of the circle, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. \bigodot{F} has secants \overline{AE} and \overline{AD} which intersects at an in point A.

Based on the theorem, it follows that \overline{AC} x \overline{CE} = \overline{AB} x \overline{BD}

• When a secant and a tangent of a circle intersect at an external point of the circle, the product of the lengths of the secant segment and its external length is equal to the length of its tangent segment. \bigodot{E} has a secant \overline{AC} and a tangent \overrightarrow{BC} that intersect at an external point C.

It follows that: \overline{AC} x {\$ \overline{AD} = (DP)'^2^'