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Similar Triangles problem.

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by AndyB » Tue Jan 17, 2012 10:08 am
Hi Friends,

Need help in solving the below problem, I have just started solving the problems so it would be of great help if you could give explanations also.

In the figure given,P is a point on AB, such that AP:PB = 4:3 PQ is parallel to AC and QD is parallel to CP.What is ratio AP:PD ???








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by LalaB » Tue Jan 17, 2012 11:07 am
let AB=7
then AP=4/7*7=4 PB=3/7*7=3

AB/AP=PB/PD OR

AB/PB=AP/PD

AP/PD=7/3

OA please
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by pemdas » Tue Jan 17, 2012 12:10 pm
Initially, consider one pair of 2 similar triangles ACP and PQD
Then consider another pair of 2 similar triangles PCB and DQB
Required ratio AP/PD is equal to PB/BD. We know that PB is 3, then PB/BD=3/x provided we set BD as x. AP=4 and AP/PD=4:y provided we set y as PD.

PB=3=x+y, PD=y=3-x, thus, 3/x=4/(3-x).

Solve for x and then find y for the required proportion of 4/y

3/x=4/(3-x), 3(3-x)=4x, 9-3x=4x, 7x=9, x=9/7 and y=3-9/7=12/7

Now proportion 4/y=4 : 12/7= 4* 7/12=7/3 is the answer
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by chieftang » Tue Jan 17, 2012 1:40 pm
I also came up with 7:3.

Looked at it like this:

Given: AP:PB = 4:3 and the parallel line info, then we know PD:DB=4:3 also.

So PD = 4/7*PB. Call AP 4 & PB 3. Then PD = 12/7

So AP:PD = 4:12/7 = 28/7:12/7 = 28:12 = 7:3
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by GMATGuruNY » Tue Jan 17, 2012 2:41 pm
Imagel]

Since AC is parallel to PQ, ∆ABC is similar to ∆PBQ.
Since AB:PB = 7:3, AC:PQ = 7:3.

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Since CP is parallel to QD, ∆ACP is similar to ∆PQD.
Since AC:PQ = 7:3, AP:PD = 7:3.
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by AndyB » Tue Jan 17, 2012 6:26 pm
LalaB wrote:let AB=7
then AP=4/7*7=4 PB=3/7*7=3

AB/AP=PB/PD OR

AB/PB=AP/PD

AP/PD=7/3

OA please

Your solution is right.
The OA is 7:3
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by ronnie1985 » Tue Jan 17, 2012 9:36 pm
Using properties of similar triangles,
AP/PB = CQ/QB
Also, CQ/QB = PD/DB=4/3
PD/DB=4/3 =>DB/PD = 3/4=> DB/PD+1 = 7/4=> (DB+PD)/PD = 7/4 = > PB/PD = 7/4
PD = (4/7)PB AND AP = (4/3)PB
AP/PD = 7/3
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by Saggii27 » Tue Jul 03, 2012 11:50 am
Since AC is parallel to PQ why it can't be triangle ACP similiar to PBQ.
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