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

In triangle A B C where A ,\ B ,\ C are acute, the distances of the orthocentre from the sides are in the proportion cosA\ :cosB\ :cosC (b) sinA\ :sinB\ :sinC secA\ :secB\ :secC (d) tanA\ :tanB\ :tanC

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

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

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 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 a Delta ABC , if p,q,r are the distances from the orthocentre to the vertices A,B,C respectively, then a q r + b r p+ c p q =

In a triangle ABC, angelA=30^(@), BC=2+ sqrt(5) , then the distance of the vertex A from the orthocentre of the triangle is

A right triangle has sides 'a' and 'b' where a>b . If the right angle is bisected then find the distance between orthocentres of the smaller triangles using coordinate geometry.

In a Delta ABC , the distance of orthocentre ( O ) from the vertices A, B , C are x,y,z respectively show that x+ y+ z = 2 ( r + R )