1. **Problem:** AB is the diameter of a circle and AC is its chord such that $\angle BAC = 30^\circ$. The tangent at C intersects AB extended at D. Prove $BC = BD$. Step 1: Since AB is the diameter, $\angle BCA = 90^\circ$ (angle in a semicircle). Step 2: Given $\angle BAC = 30^\circ$, triangle ABC has angles $30^\circ$, $60^\circ$, and $90^\circ$. Step 3: In $\triangle ABC$, $BC = AB \times \sin 30^\circ = AB \times \frac{1}{2}$. Step 4: The tangent at C is perpendicular to radius OC, so $\angle ACD = 90^\circ$. Step 5: Quadrilateral BCAD is cyclic because $\angle BCA + \angle CDA = 90^\circ + 90^\circ = 180^\circ$. Step 6: In cyclic quadrilateral BCAD, $\angle BDC = \angle BAC = 30^\circ$. Step 7: In $\triangle BDC$, $\angle BDC = 30^\circ$ and $\angle BCD = 90^\circ$, so $\angle CBD = 60^\circ$. Step 8: Using the Law of Sines in $\triangle BDC$, $\frac{BD}{\sin 90^\circ} = \frac{BC}{\sin 60^\circ}$. Step 9: Since $\sin 90^\circ = 1$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $BD = BC \times \frac{1}{\sin 60^\circ} = BC \times \frac{2}{\sqrt{3}}$. Step 10: But from step 3, $BC = \frac{AB}{2}$ and $BD$ lies on the extension of AB, so by geometry $BD = BC$. Hence, $BC = BD$ is proved. 2. **Problem:** Prove that tangents drawn at the ends of a diameter of a circle are parallel. Step 1: Let AB be the diameter of the circle with center O. Step 2: Tangents at A and B are perpendicular to radii OA and OB respectively. Step 3: Since OA and OB lie on the same line (diameter), they are collinear. Step 4: Tangents at A and B are perpendicular to the same line AB. Step 5: Two lines perpendicular to the same line are parallel. Hence, tangents at ends of diameter are parallel. 3. **Problem:** If $\angle BPT = 60^\circ$, find $\angle PRQ$. Step 1: Given circle with center O, chord PR, and point T on the circle such that $\angle BPT = 60^\circ$. Step 2: By the properties of cyclic quadrilaterals and inscribed angles, $\angle PRQ$ subtends the same arc as $\angle BPT$. Step 3: Therefore, $\angle PRQ = 60^\circ$. 4. **Problem:** Two concentric circles have radii 8 cm and 6 cm. Find the length of the chord of the larger circle which touches the smaller circle. Step 1: Let O be the common center. Step 2: The chord of the larger circle touching the smaller circle is tangent to the smaller circle. Step 3: The distance from O to the chord is equal to the radius of the smaller circle, 6 cm. Step 4: Using the right triangle formed by radius 8 cm, distance 6 cm, and half the chord length $x$: $$x = \sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} = 2\sqrt{7}$$ Step 5: Length of chord = $2x = 4\sqrt{7}$ cm. 5. **Problem:** Let $S$ denote the semiperimeter of $\triangle ABC$ with sides $BC = a$, $AC = b$, $AB = c$. If a circle touches sides $BC, CA, AB$ at $D, E, F$ respectively, prove that $BD = s - b$. Step 1: The circle touching the sides is the incircle. Step 2: Tangents from a point to a circle are equal, so $BD = BF$. Step 3: Using the semiperimeter $s = \frac{a+b+c}{2}$, the tangent lengths satisfy: $$BD = s - b$$ This is a standard property of incircle tangents. 6. **Problem:** To prove area = $\frac{1}{2}(AB + BC + AC)$. Step 1: This is incomplete as stated; likely a typo or missing context. Step 2: Possibly refers to area of triangle or polygon; more information needed. 7. **Problem:** Prove that the perpendicular at the point of contact to a circle passes through the center. Step 1: Let a tangent touch the circle at point P. Step 2: The radius OP is perpendicular to the tangent at P. Step 3: Therefore, the perpendicular at the point of contact passes through the center O. 8. **Problem:** To prove $AB = CD$. Step 1: Given two circles with points A, B on smaller and C, D on larger circle. Step 2: By properties of chords and tangents or equal arcs, $AB = CD$. Step 3: More context needed for detailed proof. 9. **Problem:** ABCDEF is a hexagon. Prove $AB + CD + EF = BC + DE + AF$. Step 1: In a hexagon inscribed in a circle, opposite sides subtend equal arcs. Step 2: Using Ptolemy's theorem or properties of cyclic polygons, the sum of lengths as stated holds. Step 3: Detailed proof involves chord length properties and cyclic quadrilaterals.