In a triangle ABC, the vertices A,B,C are at distances of p,q,r fom the orthocentre respectively. Show that ` aqr+brp+cpq=abc`

In triangle ABC where,A,B,C are acute, the distances of the orthocentre from the sides are in the proportion

A triangle ABC with vertices A(-1,0),B(-2,3/4)EC(-3,-7/6) has its orthocentre

If in a triangle ABC,a,b,c are in AP.and p_(1),p_(2),p_(3) are the altitudes from the vertices A,B,C respectively then

The co-ordinates of the vertices P,Q,R&S of square PQRS inscribed in the triangle ABC with vertices A=(0,0),B(3,0)&C=(2,1) given that two of its vertices P,Q are on the side AB are respectively:

In any !ABC ,the distance of the orthocentre from the vertices A, B,C are in the ratio

If a Delta ABC has vertices A ( - 1,7) , B (-7,1) and C(5,-5) then its orthocentre has coordinates :

If x,yandz are the distances of incenter from the vertices of the triangle ABC ,respectively, then prove that (abc)/(xyz)=cot((A)/(2))cot((B)/(2))cot((C)/(2))

If in a Delta ABC , O and O' are the incentre and orthocentre respectively, then (O' A+O'B +O'C) is equal to

If A(0,2,3) , B=(2,-1,5) are two vertices of triangle ABC whose orthocentre is H . Then distance between the orthocentres of triangle HAC and triangle HBC is