In a `triangleABC` the angles at B and C are acute. If BE and CF be drawn perpendiculars on AC and AB respectivel, prove that
`BC^(2) = Abxx BF + AC xx EC`

In a triangle A B C , the angles at B and C are acute. If BE and CF be drawn perpendiculars on AC and AB respectively, prove that B C^2=A B*B F+A C*C Edot

ABC is a triangle in which BE and CF are perpendiculars to AC and AB respectively. If BE=CF, prove that triangleABC is isosceles.

In DeltaABC, angleB and angleC are acute angles. BE and CF are perpendicular from vertices B and C on sides AC and AB respectively. Then BC^(2) is equal to-

In the isosceles triangle ABC, AB=AC and BE is perpendicular to AC from B. Prove that BC^(2) = 2ACxxCE

In DeltaABC , AD is drawn perpendicular to BC. If BD:CD=3:1 , then prove that BC^(2)=2(AB^(2)-AC^(2)) .

In triangleABC, AB = AC , then |__ B =___

In the given, if AB = AC. prove the BE = EC.

In the figure, AD is perpendicular to BC, prove that AB+AC gt 2AD .

In the figure, AD is perpendicular to BC, prove that AB+AC gt 2AD .

Delta ABC is an obtus triangle in which /_B = obtus angle. If AD is perpendicular to BC ( or extended BC ) ,then prove that AC^(2) = AB^(2) +BC^(2) + 2BC xx BD .