In a triangle ` A B C ,` the angles at `Ba n dC` are acute. If `B Ea n d C F` be drawn perpendiculars on `A Ca n dA B` respectively, prove that `B C^2=A B*B F+A C*C Edot`

Explanation

In triangle ABC, angle B is acute and CF is perpendicular AB

so, `(AC)^2 = (AB)^2 + (BC)^2 - 2(AB)(BF) `...eq(1)

In triangle ABC, angle C is acute and CF is perpendicular AB

so, `(AB)^2 = (AC)^2 + (BC)^2 - 2(AC)(CE) `...eq(2)

Adding eq(1) and eq(2), we get

`(AC)^2 + (AB)^2 = (AB)^2 + (BC)^2 - 2(AB)(BF) +(AC)^2 + (BC)^2 - 2(AC)(CE) `

`:. 2(BC)^2 - 2((AB)(BF)+(AC)(CE))=0`

`:. (BC)^2= (AB)(BF)+(AC)(CE)`

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