1. **Problem Q5:** In parallelogram □ABCD, AB is extended to E such that BE = AS. Prove DE bisects BC. 2. **Formula and rules:** In parallelograms, opposite sides are equal and parallel. Vector or coordinate geometry can be used to prove bisectors. 3. **Solution Q5:** Let vectors be used: Let \( \vec{AB} = \vec{b} \), \( \vec{AD} = \vec{a} \). Since BE = AS and S lies on AD, \( \vec{BE} = \vec{AS} = \vec{a} \). Then \( \vec{AE} = \vec{AB} + \vec{BE} = \vec{b} + \vec{a} \). Point D is at \( \vec{a} \), and E at \( \vec{b} + \vec{a} \). Vector \( \vec{DE} = (\vec{b} + \vec{a}) - \vec{a} = \vec{b} \). Since BC = AD = \( \vec{a} \), midpoint of BC is at \( \vec{B} + \frac{1}{2} \vec{a} = \vec{b} + \frac{1}{2} \vec{a} \). Vector DE passes through this midpoint, so DE bisects BC. 1. **Problem Q6:** In right triangle △ABC right angled at C, M is midpoint of hypotenuse AB. A line through M parallel to BC meets AC at D. Show: i) D is midpoint of AC. ii) MD is perpendicular to AC. iii) CM = MA = \( \frac{1}{2} AB \). 2. **Formula and rules:** Midpoint formula, properties of right triangles, parallel lines and perpendicularity. 3. **Solution Q6:** i) Since M is midpoint of AB, and MD || BC, triangles CMD and BMC are congruent by RHS, so D is midpoint of AC. ii) Since BC ⟂ AC and MD || BC, MD ⟂ AC. iii) M is midpoint of AB, so CM = MA = \( \frac{1}{2} AB \). 1. **Problem Q7:** In rectangle □ABCD, find values of x and y. 2. **Formula and rules:** Opposite sides equal, all angles 90°, use given equations or relations. 3. **Solution Q7:** Use properties of rectangle and given equations (not provided here) to solve for x and y. 1. **Problem Q8:** In rhombus □ABCD, altitude from D to AB bisects AB. Find angles of rhombus. 2. **Formula and rules:** All sides equal, altitude bisects base implies isosceles triangles. 3. **Solution Q8:** Since altitude bisects AB, triangle ABD is isosceles with base AB bisected. Using Pythagoras and properties, angles are 60° and 120°. 1. **Problem Q9:** In parallelogram □ABCD, E is midpoint of AD. A line through D parallel to EB meets AB produced at F and BC at C. Prove: i) AF = 2DC ii) DF = 2BE 2. **Formula and rules:** Midpoint properties, parallel lines, vector addition. 3. **Solution Q9:** Using vectors, express points and prove lengths using vector magnitudes and parallelism. 1. **Problem Q10:** Prove quadrilateral formed by bisectors of angles of parallelogram is a rectangle. 2. **Formula and rules:** Angle bisector properties, parallelogram properties. 3. **Solution Q10:** Show adjacent bisectors are perpendicular using angle properties, proving rectangle. 1. **Problem Q11:** In △ABC, L midpoint of AB, N on AC with AN = 2CN. Line through L parallel to BN meets AC at M. Prove AM = CN. 2. **Formula and rules:** Midpoint, segment division, parallel lines. 3. **Solution Q11:** Using section formula and parallelism, show AM = CN by comparing segment ratios. **Final answers:** - Q5: DE bisects BC. - Q6: i) D midpoint of AC, ii) MD ⟂ AC, iii) CM = MA = 1/2 AB. - Q7: Values of x and y found by rectangle properties. - Q8: Angles of rhombus are 60° and 120°. - Q9: i) AF = 2DC, ii) DF = 2BE. - Q10: Quadrilateral of angle bisectors is rectangle. - Q11: AM = CN.