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AB=a+b,
AC=a+c
B C = b + c
In Δ C D B ,
C D 2 = B C 2 − B D 2
⇒ C D 2 = ( b + c ) 2 − ( b − c ) 2
⇒ C D 2 = ( b + c + b − c ) ⋅ ( b + c − b + c ) = 2 b ⋅ 2 c = 4 b c
⇒ C D = √ 4 b c = 2 √ b c
Similarly, C F 2 = ( a + c ) 2 − ( a − c ) 2
⇒ C F = 2 √ a c ⇒ D F = C F + C D
= 2 √ a c + 2 √ b c = 2 √ c ( √ a + √ b )
In Δ A E B , B E 2 = A B 2 − A E 2 ⇒ B E 2
= ( a + b ) 2 − ( b − a ) 2 ⇒ B E 2 = ( a + b + b − a ) ⋅ ( a + b − b + a )
= 2 b ⋅ 2 a = 4 a b
⇒ B E = √ 4 a b = 2 √ a b
As B E = D F ,
⇒ 2 √ a b = 2 √ c ( √ a + √ b )
⇒ 1 √ c = √ a + √ b √ a b
⇒ 1 √ c = √ a √ a b + √ b √ a b
⇒ 1 √ c = 1 √ b + 1 √ a
⇒ 1 √ c = 1 √ a + 1 √ b
Hence Proved
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