SOLUTION: In Triangle ABC, AB=9, BC=6, and AC=12. Each of the three segments drawn though point K has length x and is parallel to a side of the triangle. Find the value of x.

Question 69363This question is from textbook
: In Triangle ABC, AB=9, BC=6, and AC=12. Each of the three segments drawn though point K has length x and is parallel to a side of the triangle. Find the value of x. This question is from textbook

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
THIS IS QUITE AN INTERSETING PROBLEM!!THANKS FOR BRINGING IT!!
In Triangle ABC, AB=9, BC=6, and AC=12. Each of the three segments drawn though point K has length x and is parallel to a side of the triangle. Find the value of x.
PLEASE MAKE A DRAWING/SKETCH AS DESCRIBED BELOW TO CLEARLY UNDERSTAND THE
PROBLEM
1.ABC IS THE TRIANGLE AS GIVEN.BC IS BASE.ANGLE ABC WILL BE OBTUSE.
2.MARK A POINT K INSIDE APPROXIMATELY SATISFYING THE GIVEN CONDITION
THAT KD=KE=KF=X,WHERE D,E,F ARE POINTS ON BC,CA,AB SUCH THAT
KD||CA....KE||AB....KF||BC
3.EXTEND KD,KE,KF TO MEET AB,BC AND CA AT F',D' AND E'RESPECTIVELY.
4.LET DC=P......EA=Q......FB=R
5.FROM PARALLELOGRAMS,WE KNOW
KF=BD'=X=KE=AF'=KD=CE'
PROOF
WE FOLLOW THE THEOREM THAT IF A LINE IS PARALLEL TO BASE OF A TRIANGLE
IT DIVIDES THE OTHER 2 SIDES IN PROPORTION..
IN TRIANGLE ABC,FE'||BC...HENCE
AF/FB=AE'/E'C
(9-R)/R = (12-X)/X......R=3X/4
SIMILARLY
BF'/F'A=BD/DC
(9-X)/X=(6-P)/P...............P=2X/3
CE/EA=CD'/D'B
(12-Q)/Q = (6-X)/X................Q=2X
NOW TRIANGLES AFE' AND ABC ARE SIMILAR.
FE'=FK+KE'=X+2X/3=5X/3
AF=9-3X/4
HENCE
FE'/BC=AF/AB
5X/(3*6)=[9-(3X/4)]/9
5X/18=1-X/12
X=36/13...
HOPE IT IS CLEAR. PLEASE CHECK BY A DRAWING TO SCALE AND COME BACK IF AIN ANY DOUBT.

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