Question

ABC is a triangle.point D is a point on BC. AB=c, AC=b, BD=u, DC=v, AD=t; then prove that b²u+c²v= (u+v) (uv+t²)

Answer 1

Answer:

Here's one approach.  This is using vectors.  I hope this is good for you.

Let A be the origin, so |B| = c, |C| = b and |D| = t.

As D divides BC in the ratio BD : DC = u : v, the vector D is given by

D = ( vB + uC ) / ( u + v )   =>  ( u + v ) D = vB + uC.

Squaring both sides (i.e. taking dot product with itself) gives

( u + v )² |D|² = ( vB + uC )²

=> ( u + v )² t² = b²u² + c²v² + 2uv B·C      ... (1)

Also,

( u + v )² = ( B - C )² = b² + c² - 2 B·C      [ this is the cosine rule ]

=>  2 B·C = b² + c² - ( u + v )²

Substituting this into equation (1) now gives

( u + v )² t² = b²u² + c²v² + uv ( b² + c² - ( u + v )² )

                = ( b²u² + c²v² + b²uv + c²uv ) - uv ( u + v )²

                = ( u + v ) ( b²u + c²v ) - uv ( u + v )²

=> ( u + v ) t² = b²u + c²v - uv ( u + v )

=> b²u + c²v = ( u + v ) ( uv + t² )