Question

pls solve it...
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Answer 1

Step-by-step explanation:

in ∆ABM and ∆BCM

(I) angleABM=angleCBM (as BM lrisects angleABC

(II) BM is the common to both traingle

(III) AB=BC (as angleA=angleV)

now,∆ABM congruent to ∆BCM ( in S-A-S case)

so, angleAMB=angleBMC ( corresponding angle of congruent traingles)

so,BM is paroendicular to AC ( proved)