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.

APBQ : APBC=4:21 and AAQP : AABC= 3:28.

Explanation: given : Here ABC is a triangle where AP:PC=1:3

let AP=1x and PC=3x, where x is the common value of AP and PC.

And AQ:QB = 3:4, similarly, let AQ=3y and QB= 4y, where y is the common value of AQ and QB.

Thus,  AC= AP+PC= 1x+3x=4x and AB= AQ+QB= 3y+4y=7y

since, we have to find out APBQ : APBC= AP×BQ: AB×PC= 1x×4y: 7y×3x=4xy:21xy=4:21

So, APBQ : APBC=4:21

Now, AAQP : AABC= AQ×AP:AB×AC=  3y×1x:7y×4x=3xy:28xy=3:28

Thus, AAQP : AABC= 3:28.

step-by-step explanation:

this is not a question

step-by-step explanation: