Proof Of Some Basic Theorem And Deductive Proofs SS1 Mathematics Lesson Note

The sum of the angles of a triangle is 180.

Given any triangle ABC

To prove: A+B+C=180°

Construction: Produce BC to a point X. Draw CP parallel to BA

Proof: Draw a triangle ABC with the following lettering:

a1=a2  (alternate angles)

b1=b2  (corresponding angles)

c+a1+b1  =  180°

C+a2+b2  =  180°

ABC  +  A  +  B =  180°

A  +  B  +  C  =  180°

RELATIONSHIP TO ANGLES ON A STRAIGHT LINE

The sum of angles on a straight line is 180°.

i.e. A + B + C = 180°

Angles on a parallel line cut by a transversal line:

The figure below is parallel lines cut by a transversal line indicating angles 1 – 8

Corresponding Angles

From the figure above, the following angles are corresponding

<1 = <5     ;      <3 = <7

<2 = <6    ;       <4 = <8

Alternate Angles

From the diagram above, the alternate angles are seen.

Vertically opposite Angles 

From the diagram above, the vertically opposite angles are:

<1 = <3        ;    <2 = <4

PARALLELOGRAM

A parallelogram is a quadrilateral which has both pairs of opposite sides parallel.

Rhombus, rectangles and squares are special examples of parallelograms. A rhombus is a parallelogram with sides of equal length.

PROPERTIES OF PARALLELOGRAM

1. The opposite sides are parallel.
2. The opposite sides are equal.

iii. The opposite angles are equal.

1. The diagonals bisect one another.

PROPERTIES OF RHOMBUS

1. All four sides are equal.
2. The opposite sides are parallel.

iii. The opposite angles are equal.

1. The diagonals bisect one another at right angles.
2. The diagonals bisect the angles.

NB: In a rectangle, all of the properties of a parallelogram are found and all four angles are right angles. In a square, all of the properties of a rhombus are found and all four angles are right angles.

INTERCEPT THEOREM

If three or more parallel lines cut off equal intercepts on a transversal, then they cut off equal intercepts on any other transversal.

Given: Three parallel lines cutting a fourth line at A, B, and C so that /AB/=/BC/ and cutting another line at X, Y, and Z respectively.

To prove:/XY/ = /YZ/.

Construction: Draw XP and YQ parallel to ABC to cut BY and CZ at P and Q respectively.

Proof:

AXPB is a parallelogram   (opp. Sides //)

XP = AB        (opp side equal)                     

Similarly  /YQ/ = /BC/        (in //gm YQCB)

/XP/ = /YQ/     (given AB = BC )

In triangles XPY, YQZ

/XP/ =/YQ/               (Proved)

X1 = x2           (corr. angles)

Y1 = y2            (corr. angles)

Therefore, triangle XPY =  triangle YQZ  (AAS)

/XY/ = /YZ/  

ASSIGNMENT 

1. In triangle ABC, <BAC= 68o and <ABC = 30o. BC is produced to X. The bisectors of <ABC and <ACX meet at P. calculate <BCP and <BPC.