If the product `(A+B)*(C+D)` is well defined,then `(A+B)*(C+D)` is equal to :
(A) `AC+AD`
(B) `AC+AD+BC+BD`
(C) `BC+BD`
(D) None of these

In Figure, O is the mid point of A B and C D . Prove that (i) A O C~= B O D (ii) AC=BD

ABC is a triangle in which AB = AC and D is any point on BC. Prove that: AB^(2)- AD^(2) = BD. CD

If A,B,C,D are four points in space, then |vec(AB)xvec(CD)+vec(BC)xxvec(AD)+vec(CA)xxvec(BD)|=k (are of /_\ABC) where k= (A) 5 (B) 4 (C) 2 (D) none of these

ABC is a triangle in which AB = AC and D is a point on AC such that BC^2 = AC xx CD .Prove that BD = BC .

D is a point on the side BC of a triangle ABC such that /_A D C=/_B A C . Show that C A^2=C B.C D .

If (b+c)/(a+d)= (bc)/(ad)=3 ((b-c)/(a-d)) then a,b,c,d are in (A) H.P. (B) G.P. (C) A.P. (D) none of these

The given figure shows an equilateral triangles ABC and D is point in AC. Prove that : (i) AD lt BD (ii) BC gt BD

If in a /_\ ABC, cosA.cosB+sinA.sinB.sinC =1,then (A) A=B (B) C=pi/2 (C) AC=BC (D) AB=sqrt(2) AC

If B is the mid point of AC and C is the mid point of BD, where A,B,C,D lie on a straight line, say why AB=CD ?

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