Question

In triangle ABC, point L and M divides the side AB and BC in the ratio 2:3 respectively.
AM and LC intersect at point P. From point P a line parallel to BA is drawn intersecting
AC at D. Find the ratio AD: DC

Answer 1

In triangle ABC, point L and M divides the side AB and BC in the ratio 2:3 respectively. AM and LC intersect at point P. From point P a line parallel to BA is drawn intersecting AC at D. Find the ratio AD: DC

Solution:

1) MK||AB||RD, (AL:LB)=(2:3), (BM:MC)=(2:3)

2) Using Menelaus' theorem we have:

$$\begin{array}{l} \frac {B M}{M C} \cdot \frac {C P}{L P} \cdot \frac {A L}{A B} = 1 \\ \frac {2}{3} \cdot \frac {C P}{L P} \cdot \frac {2}{5} = 1 \\ \frac {C P}{L P} \cdot \frac {4}{1 5} = 1 \\ \frac {C P}{L P} = \frac {1 5}{4} \end{array}$$

3) Using Thales' theorem we have:

$$\frac {L P}{C P} = \frac {A D}{C D} = \frac {4}{1 5}$$

Answer: (AD:CD)=(4:15)