Right Triangles

[karansawan]

In a right triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 1:2. Prove that

9(AQ^2 + BP^2) = 13AB^2

Please help!

[Guest]

This doesnt look like a GMAT question.
However,
if you say AP = a, then PC = 2a.
Similarly, if BQ = b, CQ = 2b.

Bigger rt. triangle ABC gives us
AB^2 = (3a)^2 + (3b)^2
= 9a^2 + 9b^2...(i)

Smaller rt triangle, PCB gives us
PB^2 = (2a)^2 + (3b)^2
= 4a^2 + 9b^2 ...(ii)
Similarly AQ^2 = 4b^2 + 9a^2...(iii)

Adding (ii) and (iii)

AQ^2 + PB^2 = 13a^2 + 13b^2
The rest follows.

In a right triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 1:2. Prove that

9(AQ^2 + BP^2) = 13AB^2

Please help!

[JonathanSchneider]

Agreed on all of the above, except that you also need to be told specifically about the ratio of 2:1. You need to know whether P is closer to A or C, and whether Q is closer to C or B. The above proof only works when P is closer to A than C, and when Q is closer to B than C.