Given: A right ∆ABC right angled at C.

To prove:
AQ² + BP² = AB² + PQ²
Proof: In right ∆ACQ, we have
AQ² = AC² + CQ²    ...(i)
[Using Pythagoras theorem]
In right ∆PCB, we have
BP² = PC² + BC²    ...(ii)
[Using Pythagoras theorem]
Adding (i) and (ii), we get
AQ² + BP² = (AC² + BC²) + (PC² + CQ²)
⇒ AQ² + BP² = AB² + PQ²    Proved.

Const: Draw AE ⊥ BC
Proof : In ∆ABE and ∆ACE, we have
AB = AC    [given]
AE = AE    [common]
and    ∠AEB = ∠AEC    [90°]
Therefore, by using RH congruent condition
∆ABE ~ ∆ACE
⇒    BE = CE
In right triangle ABE.
AB² = AE² + BE² ...(i)
[Using Pythagoras theorem]
In right triangle ADE,
AD² = AE² + DE²
[Using Pythagoras theorem]
Subtracting (ii) from (i), we get
AB² - AD² = (AE² + BE²) - (AE² + DE²)
 AB² - AD² = AE² + BE² - AE² - DE²
⇒ AB² - AD² = BE² - DE²
⇒ AB² - AD² (BE + DE) (BE - DE)
But    BE = CE    [Proved above]
⇒ AB² - AD² = (CE + DE) (BE - DE)
= CD.BD
⇒ AB² - AD² = BD.CD Hence Proved.

We know that,
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

And 

From (i) and (ii)
                
Hence,          

Let BQ = QP = PC = x
Then,    BP = 2x and BC = 3x
In ∆ABQ,
AQ² = AB² + BQ²
[Using Pythagoras theorem]
⇒    AQ² = AB² + x²    ...(i)
In ∆ABP,
AP² = AB² + BP²
[Using Pythagoras theorem]
⇒    AP² = AB² + (2x)²
⇒    AP² = AB² + 4x²    ...(ii)
In ∆ABC,
AC² = AB² + BC²
[Using Pythagoras theorem]
⇒    AC² = AB² + (3x)²
⇒    AC² = AB² + 9x²    ...(iii)
Multiplying (ii) by 8
8AP² = 8AB² + 32x² ...(iv)
Multiply (iii) by 3
3AC² = 3AB² + 21x² ...(v)
Multiply (i) by ‘5’.
5AQ² = 5AB² + 5x² ...(vi)
Adding, (v) and (vi), we get
3AC² + 5AQ² = (3AB² + 5AB²)+ (27x² + 5x²)
⇒ 3AC² + 5AQ² = 8AB² + 32x² [from (iv)]
3AC² + 5AQ² = 8AP².