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प्रश्न
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 = `"p"^2 + "a"x + "a"^2/4`
उत्तर
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`
b2 = `"p"^2 + "a"x + 1/4 "a"^2`
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