In triangle ABC, the median BM_2 intersects the bisector AL_1 at point P.
The side BC is divided by the base of the bisector AL_1 into segments CL_1=m and BL_1=n.
Determine the ratio of the segments AP to PL_1.
By Theorem of Menelaus in triangle ACL_1 we obtain (AP/PL_1)(BL_1/BC)(CM_2/AM_2)=1
Consequently AP/PL_1 = BC/BL_1 =(m+n)/n, because CM_2/AM_2=1
Then AP/PL_1=(m+n)/n, independently of the fact that AL_1 bisects the angle BAC, i.e. the above result holds for P arbitrary on BC such that CL_1=m and BL_1=n.
If moreover AL_1 bisects the angle BAC, then BL_1/CL_1=AB/AC => n/m=c/b, with AB=c and AC = b => n/m+n=c/(b+c). So in this case we can find AP/PL_1=(b+c)/c, using only the lenghts of ABC triangle.
A little more complicated solution can be given using areas.....(AM_2/AC)(AP/AL_1)=area(APM_2)/area(ACL_1) ..............
All the best Liudmyla, hope you are fine and in good health! Slava Ukraina!
... a slightly more geometric solution ...
Let N be the midpoint of CL, then CN=NL=m/2, and let LP=x.
Similarity of triangles BLP and BNM implies
BL/BN = LP/NM, which is equivalent to n/(n+m/2) = x/NM.
Hence, NM = x(2n+m)/(2n), which is a mid-segment in triangle CLA, so AL = 2*NM = x(2n+m)/n.
Next, AP = AL - x = x(2n+m)/n - x = x(n+m)/n.
Finally, AP/PL = x(n+m)/n / x = (n+m)/n.
Dinu Teodorescu Steftcho P. Dokov
Dear colleagues,
I am grateful for your interest in my geometric problem. Both of your solutions are correct and they are excellent!
I also want to express my heartfelt appreciation for your words of support for my country during these difficult times.
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