In the following figure, BL=CM.Prove that AD is a median of triangle ABC
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Answered by
56
in triangles LBD and MCD,
angle BLD=angle CMD
angle LDB= angle MDC
so, angle DBL=angle DCM
and BL=CM
then, triangle LBD congruent to triangle MCD[ASA]
so, BD=DC [C.P.C.T.]
therefore, AD is a median of triangle ABC.
angle BLD=angle CMD
angle LDB= angle MDC
so, angle DBL=angle DCM
and BL=CM
then, triangle LBD congruent to triangle MCD[ASA]
so, BD=DC [C.P.C.T.]
therefore, AD is a median of triangle ABC.
Answered by
57
in tri.BLD and tri.DMC we have
BL=CM. (given)
ang.DLB=ang.DMC. (90 each)
ang. MCD=ang. DBL
SO tri.DLB is congruent to tri.DMC by ASA theorem
so, BD=DC by C.P.C.T
SO, AD is median of tri.ABC
BL=CM. (given)
ang.DLB=ang.DMC. (90 each)
ang. MCD=ang. DBL
SO tri.DLB is congruent to tri.DMC by ASA theorem
so, BD=DC by C.P.C.T
SO, AD is median of tri.ABC
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