ABC is triangle. D is the middle point of BC. If AD is perendicular to AC, then prove that
`cosAcosC=(2(c^(2)-a^(2)))/(3ac)`

ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, The value of cos A cos C , is

In a Delta ABC, if D is the mid point of BC and AD is perpendicular to AC, then cos B=

D is midpoint of BC in Delta ABC such that AD and AC are perpendicular,Show that cos A cos C=(2(c^(2)-a^(2)))/(3ac)

If D id the mid-point of the side BC of a triangle ABC and AD is perpendicular to AC , then

In Delta ABC and D is the mid point of BC If DL perp AB and DM perp AC such that DL=DM prove that AB=AC

ABC is a triangle and D is the mid-point of BC .The perpendiculars from D to AB and AC are equal.Prove that the triangle is isosceles.

If D is the mid-point of the side BC of a triangle ABC, prove that vec AB+vec AC=2vec AD .

In a triangle ABC,AC>AB,D is the mid- point of BC and AE perp BC .Prove that: (i) AC^(2)=AD^(2)+BCDE+(1)/(4)BC^(2)

In a right triangle ABC, right angled at A, AD is drawn perpendicular to BC. Prove that: AB^(2)-BD^(2)= AC^(2)-CD^(2)