Higher June 2018 Paper 3 Q21
21
\(ABCD\) is a parallelogram.
\(ABP\) and \(QDC\) are straight lines.
Angle \(ADP\) = angle \(CBQ = 90^\circ\)
(a) Prove that triangle \(ADP\) is congruent to triangle \(CBQ\). (3)
(b) Explain why \(AQ\) is parallel to \(PC\). (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof | C1 | for starting the proof, identifying a pair of relevant equal sides or angles with reasons from \(AD = BC\) (opposite sides of a parallelogram are equal) angle \(PAD\) = angle \(QCB\) (opposite angles of a parallelogram are equal) angle \(ADP\) = angle \(CBQ\) (given or both 90°) |
| C1 | (dep C1) for complete identification of all three equal aspects with reasons | |
| C1 | (dep C2) for conclusion of congruency proof |
Additional guidance
Congruency conclusion must include a reference to ASA
| Answer | Mark | Mark scheme |
|---|---|---|
| Explanation | C1 | for identifying a pair of equal sides or angles in \(APCQ\), with reason, eg \(AP = QC\) since triangle \(ADP\) is congruent to triangle \(CBQ\) |
| C1 | (dep C1) for reasoning that \(APCQ\) is a parallelogram so opposite sides of a parallelogram are parallel |