Past Paper Example 1
GCSE Jun 2018 3H Q21
5 marks
ABCD is a parallelogram, ABP and QDC are straight lines, angle ADP = angle CBQ = 90°. (a) Prove triangle ADP is congruent to triangle CBQ (3) (b) Explain why AQ is parallel to PC (2)
ABCD is a parallelogram.
ABP and QDC are straight lines.
Angle ADP = angle CBQ = 90°
ABP and QDC are straight lines.
Angle ADP = angle CBQ = 90°
(a) Prove that triangle ADP is congruent to triangle CBQ. [3 marks]
(b) Explain why AQ is parallel to PC. [2 marks]
Worked Solution & Mark Scheme (5 marks)
Part (a) Solution:
AD = BC (opposite sides of parallelogram), ∠ADP = ∠CBQ = 90° (given), ∠PAD = ∠QCB (opposite angles of parallelogram ABCD) M2
⇒ congruent by ASA C1
⇒ congruent by ASA C1
Answer (a): AD = BC (opposite sides of parallelogram ABCD are equal); angle ADP = angle CBQ = 90° (given); angle PAD = angle QCB (alternate angles as AB is parallel to DC and AD is parallel to BC, or opposite angles of parallelogram ABCD). Therefore triangle ADP is congruent to triangle CBQ by ASA.
Part (b) Solution:
AP = QC from congruence and AP is parallel to QC, so APCQ is a parallelogram and its opposite sides AQ and PC are parallel M1 C1
Answer (b): Since triangles ADP and CBQ are congruent, AP = QC. Also AP is parallel to QC (since ABP and QDC are parallel lines). A quadrilateral APCQ with a pair of opposite sides equal and parallel is a parallelogram, so AQ is parallel to PC.