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)`

In Delta ABC , D is the mid point of BC and AD is perpendicular to AC , then show that cos A cos C = ( 2 ( c^(2) - a^(2))/( 3ac)

In triangle ABC, D is the mid point of side BC . If AD is perpendicular to AC . then prove that cosA cosC =(2(c^2-a^2))/(3ca)

ABC is a triangle and D is the middle point of BC. If AD is perpendicular to AC, then prove that cos A . cos C =(2(c^(2)-a^(2)))/(3ac)

If D is a point of the base BC an isosceles triangle ABC such that AD is perpendicular to AC, then a=

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

ABC is a triangle and D is the middle point of BC. If AD is perpendicular to AC, prove that, cosA cos C =(2(c^(2)-a^(2))/(3ca))

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

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

In Delta ABC, AD and BE are perpendiculars from A and B to the sides BC and AC, then