In ∆ABC, D, E and F are the midpoints of BC, AB and AC respectively. The lines AD and CE intersect at point K.
(a) Given that AB = 6p and AC = 10q, express in terms of p and q:
(i) BC
(ii) AD
(iii) CE
(b) Given that AK = h AD and by considering ∆ABK, express BK in terms of h, p and q.
(c) Given that CK = k CE and by considering ∆BCK, express BK in terms of k, p and q.
(d) Using these 2 expressions for BK, find the values of h and k.
(e) Prove that BK, when produced, will pass through F.
*************************
Answer:
(a)
(i) BC = BA + AC = -6p + 10q
(ii) AD = AB + BD
= AB + ½BC
= 6p + ½(-6p + 10q)
= 3p + 5q
(iii) CE = CA + AE
= CA + ½AB
= -10q + ½(6p)
= 3p - 10q
(b) BK = BA + AK
= BA + hAD
= -6p + h(3p + 5q)
= (3h - 6)p + 5hq
(c) BK = BC + CK
= BC + kCE
= -6p + 10q + k(3p - 10q)
= (3k - 6)p + (10 - 10k)q
(d) Comparing coefficients of p,
3h - 6 = 3 k - 6
h = k
Comparing coefficients of q,
5h = 10 - 10k
Sub h = k
5k = 10 - 10k
15k = 10
k = h = 2/3
(e) BF = BA + AF
= BA + ½AC
= -6p + ½(10q)
= -6p + 5q
BK = [3(⅔) - 6] p + 5(⅔)q
= -4p + (10/3) q
= ⅔(-6p + 5q)
= ⅔BF
Since BK = ⅔BF, hence, BK produced will meed AC at F.
