1. **Problem Statement:** Given a circle with center $C$, chord $MP$ with midpoint $E$, diameter $F$, and angle $\angle HCP = 130^\circ$, find the angles $\angle HMP$, $\angle HEM$, $\angle MHJ$, $\angle MPJ$, $\angle PCE$, and $\angle CPE$. Also, calculate the area of $\triangle ABD$ given $CD=5$ cm and $BC=3$ cm. 2. **Key Formulas and Rules:** - The angle subtended by a chord at the center is twice the angle subtended at the circumference on the same side. - The midpoint of a chord divides it into two equal segments. - The diameter subtends a right angle to any point on the circle. - Area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ or use Heron's formula if all sides are known. 3. **Step-by-step Solutions:** **a) Find $\angle HMP$:** - $\angle HCP = 130^\circ$ is the angle at the center subtended by arc $HP$. - The angle at the circumference subtended by the same arc $HP$ is half of $130^\circ$. - Therefore, $\angle HMP = \frac{130^\circ}{2} = 65^\circ$. **b) Find $\angle HEM$:** - $E$ is midpoint of chord $MP$, so $EM = EP$. - Triangle $HEM$ is isosceles with $EM = EP$. - Since $\angle HMP = 65^\circ$, and $E$ lies on $MP$, $\angle HEM$ is also $65^\circ$ by symmetry. **c) Find $\angle MHJ$:** - $F$ is diameter, so $\angle MHJ$ subtended by diameter is $90^\circ$. **d) Find $\angle MPJ$:** - $\angle MPJ$ is an angle subtended by chord $MJ$ at point $P$. - Since $F$ is diameter and $J$ lies on the circle, $\angle MPJ = 90^\circ$. **e) Find $\angle PCE$:** - $E$ is midpoint of chord $MP$, so $CE$ is perpendicular bisector of $MP$. - Therefore, $\angle PCE = 90^\circ$. **f) Find $\angle CPE$:** - Since $CE$ is perpendicular bisector of $MP$, $\angle CPE = 90^\circ$. **Area of $\triangle ABD$:** - Given $CD = 5$ cm (radius), $BC = 3$ cm. - $BD$ is diameter $= 2 \times CD = 10$ cm. - $\triangle ABD$ is right-angled at $B$ because $BD$ is diameter. - Use Pythagoras to find $AB$: $$AB = \sqrt{AD^2 - BD^2} = \sqrt{(BC + CD)^2 - BD^2} = \sqrt{(3 + 5)^2 - 10^2} = \sqrt{8^2 - 10^2} = \sqrt{64 - 100}$$ This is invalid (negative), so re-examine. - Actually, $BC$ and $CD$ are segments on the radius line, so $AD = BC + CD = 3 + 5 = 8$ cm. - $BD = 10$ cm (diameter). - Use Heron's formula: - Sides: $AB$, $BD=10$, $AD=8$. - Find $AB$ using Pythagoras in $\triangle ABC$ or $\triangle ABD$. - Since $BD$ is diameter, $\angle BAD = 90^\circ$. - Area: $$\text{Area} = \frac{1}{2} \times AD \times BC = \frac{1}{2} \times 8 \times 3 = 12 \text{ cm}^2$$. **Final answers:** - $\angle HMP = 65^\circ$ - $\angle HEM = 65^\circ$ - $\angle MHJ = 90^\circ$ - $\angle MPJ = 90^\circ$ - $\angle PCE = 90^\circ$ - $\angle CPE = 90^\circ$ - Area of $\triangle ABD = 12$ cm$^2$