Question

E and F are interior points on side AC and AB , respectively of triangle ABC . The line BE and CF intersect in P. If area of triangle BPF is 5, the area of the quadrilateral AEPF is 22 and area of triangle CPE is 8, then find area of triangle BPC​

Answer 1

As the figure explains, E and F are the points on AC and AB. P is the point of intersection of BE and CF.

Given:

Area of ΔBPF = 5

Area of ΔCPE = 8

Area of quadrilateral AEPF = 22

Find

ΔBPC

Solution:

Let's connect points E and F. Now the quadrilateral AEPF is divided into 2 parts. Let's call them x and y.  

For triangle EBC and EBF,

$\frac{EP}{BP} = Area \frac{CEP}{CPB}$

Similarly,  

$\frac{EP}{BP} = Area \frac{EPF}{BPF}$

equating both equations:

$Area \frac{CEP}{CPB} = \ Area \frac{EPF}{BPF}$

$\frac{8}{Area CPB} = \frac{x}{5}$

as we know, x = 22- y

placing in equation:

$\frac{8}{Area CPB} = \frac{22-y}{5}$

22CPB - y CPB  = 40................. (1)

Now looking at triangle AFE and BFE

$\frac{AF}{FB} = Area \frac{AFE}{EFB}$

and

$\frac{AF}{FB} = Area \frac{ACF}{BCF}$

equating both equations:

$Area \frac{AFE}{EFB} = \ Area \frac{ACF}{BCF}$

$\frac{y}{5} = \frac{22+8}{CBP+5}$

y (CBP+5) = 150.............(2)

equating equation 1 and 2

22CPB -yCPB  = 40

5y+ yCPB =150

Adding both equation

27 CPB = 190

CPB = 190 / 27

CPB = 7

Hence the area of BPC is 7

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