In the figure below, E is the midpoint of BC, AD: DC=3:2 and AE intersect with BD at F. Given that AB=b and AC=c, express in terms of b and c: i) AE ii) BD. If BF=tBD and AF=sAE, where t and s are scalars, find the values of t and s. What is the ratio in which F divides AE.

Huu hapa ni ufumbuzi wa swali lako:

a) i) Kueleza $\vec{AE}$ kwa kutumia $\vec{b}$ na $\vec{c}$. Step 1: Tafuta vekta $\vec{BC}$. Tunajua $\vec{BC} = \vec{AC} - \vec{AB}$. $\vec{BC} = \vec{c} - \vec{b}$ Step 2: Tafuta vekta $\vec{BE}$. Kwa kuwa E ni katikati ya BC, $\vec{BE} = \frac{1}{2}\vec{BC}$. $\vec{BE} = \frac{1}{2}(\vec{c} - \vec{b})$ Step 3: Tafuta vekta $\vec{AE}$. Tunajua $\vec{AE} = \vec{AB} + \vec{BE}$. $\vec{AE} = \vec{b} + \frac{1}{2}(\vec{c} - \vec{b})$ $\vec{AE} = \vec{b} + \frac{1}{2}\vec{c} - \frac{1}{2}\vec{b}$ $\vec{AE = \frac{1}{2}\vec{b} + \frac{1}{2}\vec{c}}$

ii) Kueleza $\vec{BD}$ kwa kutumia $\vec{b}$ na $\vec{c}$. Step 1: Tafuta vekta $\vec{AD}$. Tunapewa AD : DC = 3 : 2. Hii inamaanisha D inagawanya AC kwa uwiano wa 3:2. $\vec{AD} = \frac{3}{3+2}\vec{AC} = \frac{3}{5}\vec{AC}$ $\vec{AD} = \frac{3}{5}\vec{c}$ Step 2: Tafuta vekta $\vec{BD}$. Tunajua $\vec{BD} = \vec{AD} - \vec{AB}$. $\vec{BD} = \frac{3}{5}\vec{c} - \vec{b}$ $\vec{BD = -\vec{b} + \frac{3}{5}\vec{c}}$

b) Kutafuta thamani za $t$ na $s$. Step 1: Eleza $\vec{AF}$ kwa kutumia $s$. Tunapewa $\vec{AF} = s\vec{AE}$. Tumia $\vec{AE}$ kutoka sehemu (a)(i). $\vec{AF} = s\left(\frac{1}{2}\vec{b} + \frac{1}{2}\vec{c}\right)$ $\vec{AF} = \frac{s}{2}\vec{b} + \frac{s}{2}\vec{c} \quad (Equation 1)$ Step 2: Eleza $\vec{AF}$ kwa kutumia $t$. Tunajua $\vec{AF} = \vec{AB} + \vec{BF}$. Pia tunapewa $\vec{BF} = t\vec{BD}$. Tumia $\vec{BD}$ kutoka sehemu (a)(ii). $\vec{AF} = \vec{b} + t\vec{BD}$ $\vec{AF} = \vec{b} + t\left(-\vec{b} + \frac{3}{5}\vec{c}\right)$ $\vec{AF} = \vec{b} - t\vec{b} + \frac{3t}{5}\vec{c}$ $\vec{AF} = (1-t)\vec{b} + \frac{3t}{5}\vec{c} \quad (Equation 2)$ Step 3: Linganisha coefficients za $\vec{b}$ na $\vec{c}$ kutoka Equation 1 na Equation 2. Kulinganisha coefficients za $\vec{b}$: $\frac{s}{2} = 1-t \quad (Equation 3)$ Kulinganisha coefficients za $\vec{c}$: $\frac{s}{2} = \frac{3t}{5} \quad (Equation 4)$ Step 4: Tatua milinganyo ya wakati mmoja kupata $t$ na $s$. Kutoka Equation 3 na Equation 4, tunaweza kulinganisha upande wa kulia: $1-t = \frac{3t}{5}$ Panua pande zote mbili kwa 5: $5(1-t) = 3t$ $5 - 5t = 3t$ $5 = 8t$ $t = \frac{5}{8}$ Badilisha thamani ya $t$ katika Equation 4: $\frac{s}{2} = \frac{3}{5} \times \frac{5}{8}$ $\frac{s}{2} = \frac{3}{8}$ $s = 2 \times \frac{3}{8}$ $s = \frac{6}{8} = \frac{3}{4}$ t = \frac{5{8}, s = \frac{3}{4}}

c) Uwiano ambao F inagawanya AE. Step 1: Tumia thamani ya $s$ kutoka sehemu (b). Tunajua $\vec{AF} = s\vec{AE}$. Kutoka sehemu (b), $s = \frac{3}{4}$. $\vec{AF} = \frac{3}{4}\vec{AE}$ Hii inamaanisha kwamba vekta $\vec{AF}$ ni $\frac{3}{4}$ ya vekta $\vec{AE}$. Step 2: Tafuta uwiano wa AF : FE. Ikiwa AF ni sehemu 3 kati ya sehemu 4 za AE, basi FE ni sehemu iliyobaki. $\vec{FE} = \vec{AE} - \vec{AF} = \vec{AE} - \frac{3}{4}\vec{AE} = \frac{1}{4}\vec{AE}$ Kwa hiyo, uwiano wa AF : FE ni: $AF : FE = \frac{3}{4}\vec{AE} : \frac{1}{4}\vec{AE} = 3 : 1$ $AF : FE = 3 : 1$