Q. 305.0( 1 Vote )

# 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(i) b2 = p2 + ax + a2/4(ii)(b2+c2)=2p2 + 1/2 a2(iii) (b2 —c2) = 2ax Given: D is the mid-point of side BC and AE BC

and BC = a, AC = b, AB= c, ED = x, AD =p and AE = h

To Prove: (i) or Proof: In right triangle ∆AEC, using Pythagoras theorem, we have

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

(AE)2 + (EC)2 = (AC)2

(AE)2 + (ED + DC)2 = (AC)2

(AE)2 + (ED)2 + (DC)2 + 2 (ED)(DC) = (AC)2

[ (a + b)2 = a2 + b2 +2ab]

(AE2 + ED2) + (DC)2 + 2 (ED)(DC) = (AC)2

(AD)2 + (DC)2 + 2 (ED)(DC) = (AC)2

[ In right angled ∆AED, AE2 + ED2 =AD2] [ D is the midpoint of BC, so 2DC = BC] …(i) To Prove: (ii) or Proof: In right triangle ∆AEB, using Pythagoras theorem, we have

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

(AE)2 + (BE)2 = (AB)2

(AE)2 + (BD – ED)2 = (AB)2

(AE)2 + (ED)2 + (BD)2 – 2 (ED)(BD) = (AB)2

[ (a – b)2 = a2 + b2 – 2ab]

(AE2 + ED2) + (BD)2 – 2 (ED)(BD) = (AB)2

(AD)2 + (BD)2 – 2 (ED)(BD) = (AB)2

[ In right angled ∆AED, AE2 + ED2 =AD2] [ D is the midpoint of BC, so 2DC = BC] …(ii)

On adding eq. (i) and (ii), we get   To Prove: (iii) (b2 —c2) = 2ax

or (AC)2 – (AB)2 = 2 (BC)(ED)

Proof: Subtracting Eq. (ii) from (i), we get (AC)2 – (AB)2 = 2 (BC)(ED)

Hence Proved

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