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

Given that XY || PR and MN | QR, we can use the proportional segments theorem. From the ratio provided, we know:

$\frac{PM}{PN} = \frac{PQ}{QR}$

Here, we substitute the known values:

$\frac{PM}{PN} = \frac{35}{PQ}$

We solve for PM:

$PN = \frac{5}{7} \cdot PM$ Thus, substituting PQ = 35:

( PM = 25 \text{ units} )

Using the same properties from above, we can find XM using:

$\frac{PX}{PY} = \frac{PQ}{R}$

Substituting the known values from earlier calculations:

$PX = 35 - PM = 35 - 25 = 10 \text{ units}$ We established the total length of PQ as:

$PX = \frac{4}{35} \cdot R = 10 \text{ units}$

Therefore, solving gives: ( XM = 16.25 \text{ units} )

Using the properties of cyclic quadrilaterals, we know that the opposite angles sum to 180°. Hence, since ( ∠A = 68° ) and ( ∠B = 44° ):

• The angles equal to 44° would be:
  • ∠C = 44° (alternate angle)
  • ∠D = 44° (opposite angle in the cyclic quadrilateral).

Given that ∠C2 is on a straight line with ∠B, we use:

$∠C2 = 180° - ∠B$

Substituting:

$∠C2 = 180° - 44° = 136°.$

Since angles ∠ABD and ∠CED are alternate angles formed by a transversal cutting through the parallel lines BC and AD, we can use the property of alternate angles:

Since we have already established equal angle measures from earlier steps,

( ∠A + ∠D = 180° )

This leads to:

Thus, ∠ABD || ∠CED as proven by the properties of parallel lines.