Question

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) c² = p² - ax + a²/4
(iii) b² + c² = 2p² + a²/2

Answer 1

SOLUTION :  

Given : D is the mid-point of side BC and AE ⊥ BC. BC = a, AC = b, AB = c, ED = x, AD = p and AE = h.

BD = DC = a/2 (D is the mid-point of side BC)...........(1)

(i) In ∆ AED  

AD² = AE² + ED²

[By using pythagoras theorem]

p² = h² + x² …………….(2)

In ∆ AEC,  

AC² = AE² + EC²

[By using pythagoras theorem]

AC² = AE² +( ED + DC)²

b² = h² + ED² + DC² + 2ED× DC

[(a+b)² = a²+b² +2ab]

b² = h² +x²+ (a/2)² + 2×x×a/2

b² = h² + x²+ a²/4+ xa

b² = p²+ a²/4+ xa ……………..(3)

[From eq 2]

(ii) BD = BE + ED

BE = BD - ED

BE = a/2 - x

[From eq 1]

In ∆AEB, by pythagoras theorem

AB² = BE² + AE²

[By using pythagoras theorem]

c² = (a/2 - x)² + h²

c² = [(a/2)² +x² - 2× a/2 × x ) + h²]

[(a- b)² = a²+b² - 2ab]

c² = (a²/4 +x² - ax) + h²

c² =  h² + x² - ax +  a²/4  

c² = p² - ax +  a²/4 …………..(4)

[From eq 2]

(iii) On Adding eq 3 & 4

b² + c² = p²+ a²/4+ xa +  p² - ax +  a²/4

b² + c² = p² + p² + a²/4 + a²/4

b² + c² = 2p² + 2(a²/4)

b² + c² = 2p² + a²/2

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Answer 2

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