" (ii) "/_BAC=90;AD perp BC;PT.BC^(2)=AD*AB

In figure,/_B<90^(@) and AD perp BC .Prove that AC^(2)=AB^(2)+BC^(2)-2BC.BD

In triangle ABC, if AD perp BC and AD^(2)=BD*DC. Then find the angle BAC

If ABC is an isosceles triangle and D is a point on BC such that AD perp BC, then AB^(2)-AD^(2)=BDdot DC( b) AB^(2)-AD^(2)=BD^(2)-DC^(2)(c)AB^(2)+AD^(2)=BDdot DC(d)AB^(2)+AD^(2)=BD^(2)-DC^(2)

In an equilateral triangle ABC, if AD perp BC, then 2AB^(2)=3AD^(2)( b) 4AB^(2)=3AD^(2) (c) 3AB^(2)=4AD^(2)( d ) 3AB^(2)=2AD^(2)

In a right-angled triangle,the square of hypotenuse is equal to the sum of the squares of the two sides.Given that /_B of /_ABC is an acute angle and AD perp BC .Prove that AC^(2)=AB^(2)+BC^(2)-2BC.BD

In Fig.4.94,/_CAB=90o and AD perp BC. If AC=75cm,AB=1m and BD=1.25m find AD.( FIGURE)

In a ABC,AD perp BC and AD^(2)=BD xx CD prove that ABC is a right triangle.

In a ABC, AD perp BC and AD^(2)=BD xx CD .Prove that ABC is a right triangle.

(Result on acute triangle) In Fig.4.184,/_B of ABC is an acute angle and AD perp BC, prove that AC^(2)=AB^(2)+BC^(2)-2BC xx BD

In ABC, if AD perp BC and AD^(2)=BD xx DC, prove that /_BAC=90o