Question

In △ABC, point P∈ AC with AP:PC=1:3, point Q∈ AB so that AQ:QB=3:4, Find the ratios APBQ : APBC and AAQP : AABC.

Answer 1

1. If AP : PC = 1 : 3, then AP = x, PC = 3x.

2. If AQ : QB = 3 : 4, then AQ = 3y, QB = 4y.

3. Find the area of triangles AQP and ABC:

$A_{AQP}=\dfrac{1}{2}\cdot AP\cdot AQ\cdot \sin\angle A=\dfrac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A;$

$A_{ABC}=\dfrac{1}{2}\cdot AC\cdot AB\cdot \sin\angle A=\dfrac{1}{2}\cdot (x+3x)\cdot (3y+4y)\cdot \sin\angle A=\dfrac{1}{2}\cdot 4x\cdot 7y\cdot \sin\angleA.$

Then

$\dfrac{A_{APQ}}{A_{ABC}}=\dfrac{\frac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A}{\frac{1}{2}\cdot 4x\cdot 7y\cdot \sin\angleA}=\dfrac{3}{28}.$

4. Note that

$\dfrac{A_{APQ}}{A_{ABP}}=\dfrac{\frac{1}{2}\cdot x\cdot 3y\cdot \sin\angle A}{\frac{1}{2}\cdot x\cdot (3y+4y)\cdot \sin\angle A}=\dfrac{3}{7}.$

From the previous two ratios you have that:

$A_{ABP}=\dfrac{7}{3}A_{APQ}\Rightarrow A_{PBQ}=A_{APB}-A_{APQ}=\dfrac{7}{3}A_{APQ}-A_{APQ}=\dfrac{4}{3}A_{APQ}$

$A_{ABC}=\dfrac{28}{3}A_{APQ}\Rightarrow A_{PBC}=A_{ABC}-A_{APB}=\dfrac{28}{3}A_{APQ}-\dfrac{7}{3}A_{APQ}=7A_{APQ}.$

Then

$\dfrac{A_{PBQ}}{A_{PBC}}=\dfrac{\frac{4}{3}A_{APQ}}{7A_{APQ}}=\dfrac{4}{21}.$