Given: ΔABC is a right triangle right angled at C
P and Q are the mid–points of the sides CA and CB respectively.
⇒ AP = PC and CQ = QB
In ΔACB, using Pythagoras Theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (BC)2 = (AB)2 …(i)
Now, In ΔACQ, using Pythagoras Theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (CQ)2 = (AQ)2
⇒ 4(AC)2 + (BC)2 = 4(AQ)2
⇒ (BC)2 = 4(AQ)2 – 4(AC)2 …(ii)
Now, In ΔPCB, using Pythagoras Theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (PC)2 + (BC)2 = (BP)2
⇒ (AC)2 + 4(BC)2 = 4(BP)2
⇒ (AC)2 = 4(BP)2 – 4(BC)2 …(ii)
Putting the value of (AC)2 and (BC)2 in eq. (i), we get
4(BP)2 – 4(BC)2 + 4(AQ)2 – 4(AC)2 = (AB)2
⇒ 4(BP2 +AQ2) – 4(BC2 + AC2) = (AB)2
⇒ 4(BP2 +AQ2) – 4(AB2) = (AB)2 [from eq(i)]
⇒ 4(BP2 +AQ2) = 5(AB)2
Hence Proved
