EXERCISE 1
a In the triangles ABM and CDM :
1. angleBAM = angleDCM (alternate angles, AB || DC )
2. angleABM = angleCDM (alternate angles, AB || DC )
3. AB = CD (opposite sides of parallelogram ABCD)
triangleABM = triangleCDM (AAS)
b Hence AM = CM and DM = BM (matching sides of congruent triangles)
EXERCISE 2
From the diagram, 2α + 2β = 360o (angle sum of quadrilateral ABCD)
α + β = 180o
Hence AB || DC (co-interior angles are supplementary)
and AD || BC (co-interior angles are supplementary).
EXERCISE 3
First show that triangleABC ≡ triangleCDA using the SSS congruence test.
Hence angleACB = angleCAD and angleCAB = angleACD (matching angles of congruent triangles)
so AD || BC and AB || DC (alternate angles are equal.)
EXERCISE 4
First prove that triangleABD ≡ triangleCDB using the SAS congruence test.
Hence angleADB = angleCBD (matching angles of congruent triangles)
so AD || BC (alternate angles are equal.)
EXERCISE 5
First prove that triangleABM ≡ triangleCDM using the SAS congruence test.
Hence AB = CD (matching sides of congruent triangles)
Also angleABM = angleCDM (matching angles of congruent triangles)
so AB || DC (alternate angles are equal):
Hence ABCD is a parallelogram, because one pair of opposite sides are equal and parallel.
EXERCISE 6
Join AM. With centre M, draw an arc with radius AM that meets AM produced at C . Then ABCD is a parallelogram because its diagonals bisect each other.
EXERCISE 7
The square on each diagonal is the sum of the squares on any two adjacent sides. Since opposite sides are equal in length, the squares on both diagonals are the same.
EXERCISE 8
a We have already proven that a quadrilateral whose diagonals bisect each other is a parallelogram.
b Because ABCD is a parallelogram, its opposite sides are equal.
Hence triangleABC ≡ triangleDCB (SSS)
so angleABC = angleDCB (matching angles of congruent triangles).
But angleABC + angleDCB = 180o (co-interior angles, AB || DC )
so angleABC = angleDCB = 90o .
Hence ABCD is rectangle, because it is a parallelogram with one right angle.
EXERCISE 9
angleADM = α (base angles of isosceles triangleADM )
and angleABM = β (base angles of isosceles triangleABM ),
so 2α + 2β = 180o (angle sum of triangleABD)
α + β = 90o.
Hence angleA is a right angle, and similarly, angleB, angleC and angleD are right angles.