9.1 Complete the following theorem statement: A line drawn parallel to one side of a triangle .. - NSC Technical Mathematics - Question 9 - 2022 - Paper 2

In ( \triangle PQR ) with line segment XY drawn parallel to side PR, we can apply the basic proportionality theorem: If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. Thus, we have: [ \frac{PM}{PN} = \frac{MQ}{NR} ]\n[ PM = \frac{5}{2} \cdot NR ]\nGiven that ( PQ = 35 ) units and ( PN:NR = 5:2 ), we find: [ PQ = PM + PN = PM + \frac{2}{5} PM = \frac{7}{5} PM ]\n[ 35 = \frac{7}{5} PM \Rightarrow PM = 25 \text{ units} ]

From the previous part, we know that: ( PX = PM + MX ) and that ( PQ = PX + QY ). Given that ( PQ = 35 ) units, and we established that ( PM = 25 ), we can express MX as: [ PX + YQ = 35 \Rightarrow PX + 3YR = 35 ]\nThe calculation yields: [ PX = PM - MX = 25 - MX ]\nIn this context, we also have the ratio ( \frac{PX}{RY} = \frac{35}{4} ). On solving, this results in: [ XM = 16.25 \text{ units} ]

(\angle B_2 = 44^{\circ}) (alternate segment theorem) and (\angle C_1 = 44^{\circ}) (by inscribed angles subtended by the same arc).

Since ABCD is a cyclic quadrilateral, the opposite angles are supplementary: [ \angle A + \angle C_2 = 180^{\circ} ]\nSubstituting the known values, we have: [ 68^{\circ} + \angle C_2 = 180^{\circ} \Rightarrow \angle C_2 = 112^{\circ} ]

To prove that triangles are parallel, we show that corresponding angles are equal. Since (\angle ADB = \angle CDB) (alternate angles as line BC is parallel to AD) and given that (\angle A = \angle C = 68^{\circ}), we conclude that: [(\triangle AABD \parallel \triangle CED) \text{ by AA similarity criterion.}]