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Question
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 b2 + c2 = `2"p"^2 + "a"^2/2`
Solution
Given ∠AED = 90°
ED = x, DC = `"a"/2` ...(D is the mid point of BC)
∴ EC = `x + "a"/2`, BE = `"a"/2 - x`
∴ In the right ∆AED
AD2 = AE2 + ED2
p2 = h2 + x2
In the right ∆AEC,
AC2 = AE2 + EC2
b2 = `"h"^2 + (x + "a"/2)^2`
= `"h"^2 + x^2 + "a"^2/4 + 2 xx x xx "a"/2`
b2 = `"p"^2 + "a"^2/4 + "a"x` ......(1)
b2 = `"p"^2 + "a"x + 1/4 "a"^2` ....(2)
In the right triangle ABE,
AB2 = AE2 + BE2
c2 = h2 + (a2 – x)2
c2 = h2 + a24 + x2 – ax
c2 = h2 + x2 + 14 a2 – ax
c2 = p2 – ax + `a^2/4` .....(3)
By adding (2) and (3)
b2 + c2 = `"p"^2 + "a"x + "a"^2/4 + "p"^2 - "a"x + "a"^2/4`
= `2"p"^2 + (2"a"^2)/4`
= `2"p"^2 + "a"^2/2`
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