Bisectors of angle B and angle C in triangle ABC meet each other at X . line AX cuts the side BC in Y . Prove that AX/AY = AB+AC/BC
Answers
Answered by
6
A△XBC=rBC/2A
A△ABC=r(BC+AB+CA)/2
A△ABC/A△XBC=r(BC+AB+CA)/rBC=(BC+AB+CA)/ BC
But areas of triangles with same base are proportional to heights even if they are slant.
A△ABC/A△XBC=AY/XY=AX+XY/XY
Combining
AX+XY/XY=(BC+AB+CA)/BC
By dividendo
AX/XY=(AB+CA)/BC
A△ABC=r(BC+AB+CA)/2
A△ABC/A△XBC=r(BC+AB+CA)/rBC=(BC+AB+CA)/ BC
But areas of triangles with same base are proportional to heights even if they are slant.
A△ABC/A△XBC=AY/XY=AX+XY/XY
Combining
AX+XY/XY=(BC+AB+CA)/BC
By dividendo
AX/XY=(AB+CA)/BC
Similar questions