Q31 of 95 Page 5

P and Q are the mid–points of the sides CA and CB respectively of ΔABC right angled at C. Prove that 4(AQ2 +BP2) = 5AB2


Given: ABC ia 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 ACB, using Pythagoras Theorem, we have


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (AC)2 + (BC)2 = (AB)2 …(i)


Now, In ACQ, 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 PCB, 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


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In the given figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB= c, ED = x, AD =p and AE = h, prove that

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