Activity 1

  1. Determine the value of x, in the diagram alongside, if PQ ∣∣ BC. (4)
    Solution
    AP = AQ (PQ ∣∣ BC, prop theorem) 
    PB    QC
    5/3 = 4/x
    ∴ 5x = (3)(4)
    ∴ x = 12/5 = 2,4 cm 3 [4]
  2. In ∆ABC, AB ∣∣ FD; AF ∣∣ DE and FE : EC = 3 : 4.
    Determine EC : BF (7) 
    6
    NOTE:
    3 : 4 does not mean that
    FE = 3 and EC = 4.
    For any a, we can say that
    FE = 3a and EC = 4a
    For every 3 of a in FE, there is
    4 of a in EC.

    Solution
    Work with two different triangles:
    ∆ACF and  ABC
    7
    In ∆ACF:
    AD = FE (AF ∣∣ DE, prop intercept theorem) 
    DC   EC
    In ∆ABC:
    AD = BF (AB ∣∣ FD, prop intercept theorem)
    DC    FC
    FE = BF(both = AD)
       EC   FC             DC 
    FE = 3a and BF = BF 
    EC    4a       FC     7a
    3a = BF
       4a    7a
    ∴ BF = 3(7a/4) = 21a____4
    EC   = 4a ÷ 21a/4
       BF
    = 4a ×  4  
        1    21a
    = 16
        21
    ∴ EC : BF = 16 : 21 [7] 
  3. Determine the value of x if PQ ∣∣ BC. (4)

    Solution
    AP = AQ (prop theorem , PQ ∣∣ BC)
    PB    QC
    5/3 =  4/x
    5x = (3) (4)
    x = 12/5 = 2,4cm 
    8
    [4] 
  4. In the diagram, RF ∣∣ KG , ED ∣∣ KH,

    RH = 3 units, RK = 9 units, HF = 2 units. GE: EK = 1:3
    9
    Calculate (stating reasons) the lengths of:
    4.1 FG
    4.2 FD (8)

    Solutions
    4.1
    In ∆HKG
    FG = 9 S (line ∣∣ one side of a ∆) 3 R or (RF ∣∣ KG)
      2     3
    FG = 6 units S (3)
    4.2 GD = GE = ¼ S (line ∣∣ one side of a ∆) 3 R or (ED ∣∣ KH)
          GH    GK    
    GD =  ¼ .GH
    GD = ¼ .(8) S
    GD = 2 S
     ∴ FD = 6 − 2 = 4units 3 R
    OR
    In ∆HKG, HK ∣∣ DE
    GD = EG = 1/3 S
    DH    EK
    (line ∣∣ one side of a ∆) R or (proportional theorem, HK//DE )
    6 − FD = 1/3 S
    2 + FD
    18 − 3FD = 2 + FD
    ∴ FD = 4 units (5)
    [8] 

Activity 2

  1. Diameter AME of circle with centre M bisects FAB.
    MD is perpendicular to the chord AB.
    ED produced meets the circle at C, and CB is joined.
    16
    1. Prove ∆AEF ||| ∆AMD (5)
    2. Hence, find the numerical value of AF. (5)
                                                             AD 
    3. Prove ∆ CDB ||| ∆ADE (4)
    4. Prove AD2 = CD. DE (3)
      [17]

      Solution

      a)
      17
      F = 90° (∠ in semi-circle) 
      ^D1 = 90° (given MD ⊥ AB) 
      ∴ ^F = ^D1
      In ∆AEF and ∆AMD
      ^F = ^D1  (proved)
      ^A1 = ^A2 (AM bisects FAB) 
      ∴ ^E1 = ^M1 (third ∠ of ∆)
      ∴ ∆AEF ∣∣∣ ∆AMD (AAA) or ∠∠∠ (5) 

       Solution
      b) AE = EF = AF (||| ∆s)
          AM    MD   AD
      AM = ME (radii)
      ∴ AE = 2AM
      2AM = AF
          AM     AD
      AF = 2 (5)
         AD
      c)
      18
      In ∆CDB and ∆ADE
      ^C = ^A2 (∠s in same seg) 
      ^B = ^E2 (∠s in same seg)
      ^D4 = ^D1 + ^D2 (opp ∠)
      ∴ ∆CDB ||| ∆ADE (AAA) (4)
      Solution
      d)
      19
      CD = DB  (III ∆s)
      AD     DE
      ∴ CD.DE = AD.DB 
      But AD = DB (MD ⊥ AB, M is centre) 
      ∴ CD. DE = AD 
      ∴ AD2 = CD.DE (3)
      [17]
  2. CD is a tangent to circle ABDEF at D.
    Chord AB is produced to C. Chord BE cuts chord AD in H and chord FD in G. AC ∣∣ FD and FE = AB
    20
    1. Prove that ^D4 = ^D2 (3)
    2. Prove that ∆BHD ||| ∆FED (5)
    3. Hence AB = FD (3)
                 BH    BD

      Solutions
      a) ^A = ^D4 (tan-chord thm)
      ^D2 = ^A (alt ∠s CA ∣∣ DF)
      ^D4 = ^D2 (3)
      21

      b) In ∆BHD and ∆FED
      ^B2 = ^F(∠s in same seg) 
      ^D3 = ^D1 (equal chords) 
      ^H2 = ^E2 (third ∠ of Δ)
      ∴ ∆BHD ||| ∆FED ∠∠∠ (5)

      c) FE = FD (||| ∆s)
         BH    BD
      But FE = AB (given)
      AB = FD (3)
         BH    BD
      22
      [11] 

      [11]
  3. In the diagram ∆ABC is such that F is on AB and G is on AC. CB is produced to meet GF produced at E .DGFE is a straight line. BFA ∣∣ CD.
    AB = 20, BC = 10, EF = 8, EB = 5 and FB = 6.
    23
    3.1 Determine the numerical value of EF (3)
                                                               ED
    3.2 Calculate the length of ED (2)
    3.3 Complete, without stating the reasons: ∆EFB III ∆ …. (1)
    3.4 Hence, calculate the length of DC (3)
    3.5 Prove that: AF = FG(4)
                            CD    DG 
    [13]

    Solutions
    BFA ∣∣ CD. AB = 20, BC = 10, EF = 8, EB = 5 and FB = 6
    3.1 FB ll CD (Given) EF = EB S (line ∣∣ one side of A) 3 R
                                     ED    EC
    EF = 5 =S (3)
    ED   15   3
    3.2 EF = 1 from 3.1 and EF = 8
          ED   3
    =
      ED   3
    ED = 24 S (2)
    3.3 ∆EFB ∣∣∣ ∆EDC (1)
    3.4 DC = ED (∆EFB ∣∣∣ ∆EDC) R
          FB     EF
    DC = 24 S
     6       8
    DC = 18 S (3)
    3.5 In ∆AFG and ∆CDG
    ^A = ^C1 (alt ∠s. AF ∣∣ DC) S/R
    ^G3 = ^G1 (vertically opp ∠s) S/R
    ^F = ^D (alt ∠s. AF ∣∣ DC)
    ∆ AFG III ∆CDG (∠∠∠) R
    AF = FG (∆AFG ∣∣∣ ∆CDG) 3 R (4)
    CD    DG
     [13] 

  4. In the diagram, PQCB is a cyclic quadrilateral. Chords BP and CQ are produced to meet at A such that AQ = BC.
    24
    4.1 Prove that: ΔAPQ III ΔACB (4)
    4.2 Hence, prove that AQ² = AB.PQ (3)
    [7]
    Solutions
    4.1 Proof: In ΔAPQ and ΔACB
    ^A = ^A (common) S/R
    ^P2 = ^C S (ext ∠ of a cyclic quad ) 3 R
    ^P2 = ^ B (sum ∠s of ∆) or ( ext ∠ of cyclic quad )
    ∆APQ ∣∣∣ ∆ACB ( ∠.∠.∠) R (4)
    4.2 AQ = PQ S (∆APQ ∣∣∣ ∆ACB) S
          AB    BC 
    AQ = PQ S ( AQ = BC )
    AB     AQ 
    AQ2 = AB.PQ (3)
    [7] 
Last modified on Thursday, 02 September 2021 09:06