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.
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.
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.
In the given Fig., ∠ACB = 90° and CD ⊥ AB. Prove that
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,