In triangle ABC, AL, BM, and CN are concurrent at point P. Points U, V, and W are chosen on
AB, AC, and BC, respectively, so that LU is parallel to AC, NV is parallel to BC, and MW is
parallel to AB. Prove that AW, BV, and CU are concurrent (at point K)
Solution
From Ceva Theorem, since AL,BM,CN are concurrent, we have
BL/CL*CM/AM*AN/BN=1
LU//AC implies BL/CL=BU/AU
NV/BC implies AN/BN=AV/CV
MW//AB implies CM/AM=CW/BW
So from all four above equations we get
BU/AU*CW/BW*AV/CV=1
which again by Ceva Theorem we get AW,BV,CU are concurrent.
Measures of Central Tendency: Mean, Median and Mode
In triangle ABC, AL, BM, and CN are concurrent at point P. Points U,.pdf
1. In triangle ABC, AL, BM, and CN are concurrent at point P. Points U, V, and W are chosen on
AB, AC, and BC, respectively, so that LU is parallel to AC, NV is parallel to BC, and MW is
parallel to AB. Prove that AW, BV, and CU are concurrent (at point K)
Solution
From Ceva Theorem, since AL,BM,CN are concurrent, we have
BL/CL*CM/AM*AN/BN=1
LU//AC implies BL/CL=BU/AU
NV/BC implies AN/BN=AV/CV
MW//AB implies CM/AM=CW/BW
So from all four above equations we get
BU/AU*CW/BW*AV/CV=1
which again by Ceva Theorem we get AW,BV,CU are concurrent
