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)=(1)/(2)\cdot AP\cdot AQ\cdot \sin\angle A=(1)/(2)\cdot x\cdot 3y\cdot \sin\angle A;`

`A_(ABC)=(1)/(2)\cdot AC\cdot AB\cdot \sin\angle A=(1)/(2)\cdot (x+3x)\cdot (3y+4y)\cdot \sin\angle A=(1)/(2)\cdot 4x\cdot 7y\cdot \sin\angleA.`

Then

`(A_(APQ))/(A_(ABC))=((1)/(2)\cdot x\cdot 3y\cdot \sin\angle A)/((1)/(2)\cdot 4x\cdot 7y\cdot \sin\angleA)=(3)/(28).`

4. Note that

`(A_(APQ))/(A_(ABP))=((1)/(2)\cdot x\cdot 3y\cdot \sin\angle A)/((1)/(2)\cdot x\cdot (3y+4y)\cdot \sin\angle A)=(3)/(7).`

From the previous two ratios you have that:

`A_(ABP)=(7)/(3)A_(APQ)\Rightarrow A_(PBQ)=A_(APB)-A_(APQ)=(7)/(3)A_(APQ)-A_(APQ)=(4)/(3)A_(APQ)`

`A_(ABC)=(28)/(3)A_(APQ)\Rightarrow A_(PBC)=A_(ABC)-A_(APB)=(28)/(3)A_(APQ)-(7)/(3)A_(APQ)=7A_(APQ).`

Then

`(A_(PBQ))/(A_(PBC))=((4)/(3)A_(APQ))/(7A_(APQ))=(4)/(21).`