# 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.

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 Click here to see ALL problems on Triangles 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.