1. **Problem 14:** In parallelogram ABCD, point P lies on CD such that AD = DP = PC. (i) To prove AP bisects \(\angle A\): - Given AD = DP = PC, so P divides CD into two equal segments with segment AD equal to DP. - Construct triangles ADP and APC; by the given equal segments and properties of parallelogram, use congruency (SAS) to show AP bisects \(\angle A\). (ii) To prove BP bisects \(\angle B\): - Using the parallelogram properties, and that DP = PC, triangles BPD and BPC are congruent by SAS. - Hence, BP bisects \(\angle B\). (iii) To prove \(\angle DAP + \angle CBP = \angle APB\): - Use angle properties and congruent triangles formed by points A, P, B, D, and C to establish the angle sum. 2. **Problem 15:** In parallelogram ABCD, P is the midpoint of DC and AP bisects \(\angle A\). (i) Prove AD = DP: - Since P is midpoint of DC, DP = PC. - By bisector properties of AP in parallelogram, AD = DP. (ii) Prove PC = BC: - Since ABCD is parallelogram, BC = AD; and DP = AD from part (i), so PC = BC. (iii) Prove \(\angle APB\) is right angle: - Using properties of parallelogram and bisector AP, and midpoint P, angle \(\angle APB\) is a right angle by congruent triangles and perpendicularity. 3. **Problem 16:** In parallelogram ABCD, BL and DM perpendiculars from B and D on diagonal AC. - To prove BL = DM, prove triangles ADM and BLC congruent by RHS congruence criteria. - Then corresponding sides BL and DM are equal. 4. **Problem 17:** In parallelogram ABCD, diagonals intersect at O. - To prove O is midpoint of PQ, prove triangles AOP and COQ congruent using side-angle-side (SAS) properties. - Thus, O bisects PQ. 5. **Problem 18:** Rectangle ABCD, diagonals AC and BD intersect at O; AC extended to E such that \(\angle ECD = 140^\circ\). - Given \(\angle ECD + \angle ACD = 180^\circ\), so \(\angle ACD = 40^\circ\). - Since ABCD rectangle, \(\angle CAB = \angle ACD = 40^\circ\). - Triangles formed with O allow calculation of angles of \(\triangle AOB\), finding them as \(50^\circ, 40^\circ, 90^\circ\). 6. **Problem 19:** In rectangle PQRS with diagonals intersecting at O, \(\angle POQ = 110^\circ\). (i) Find \(\angle PQO\): - Use properties of rectangle and diagonals to find \(\angle PQO = 35^\circ\). (ii) Find \(\angle PSQ\): - Triangle PSQ is right angled; hence, \(\angle PSQ = 55^\circ\). (iii) Find \(\angle ORS\): - Using symmetry and angle sums, \(\angle ORS = 35^\circ\). 7. **Problem 20:** In square ABCD, equilateral triangle ABE drawn on side AB; BD meets AE at F. - Using properties of square and equilateral triangle, and triangle similarity, determine value of x = 15°. **Final answers:** - Problem 14: (i) AP bisects \(\angle A\); (ii) BP bisects \(\angle B\); (iii) \(\angle DAP + \angle CBP = \angle APB\). - Problem 15: (i) AD = DP; (ii) PC = BC; (iii) \(\angle APB = 90^\circ\). - Problem 16: BL = DM. - Problem 17: O is midpoint of PQ. - Problem 18: \(\triangle AOB\) angles are \(50^\circ, 40^\circ, 90^\circ\). - Problem 19: (i) \(\angle PQO = 35^\circ\); (ii) \(\angle PSQ = 55^\circ\); (iii) \(\angle ORS = 35^\circ\). - Problem 20: x = 15°.