# P, Q and R are the mid-points of sides BC, CA and AB respectively of ∆ABC. Prove that : AC + 2BQ > AB + BC

$In\u25b3ABQ,\phantom{\rule{0ex}{0ex}}AQ+BQAB\left(Sumofanytwosidesoftriangleisgreaterthanthirdside\right)\phantom{\rule{0ex}{0ex}}In\u25b3CBQ,\phantom{\rule{0ex}{0ex}}CQ+BQBC\left(Sumofanytwosidesoftriangleisgreaterthanthirdside\right)\phantom{\rule{0ex}{0ex}}Thus,AQ+BQ+CQ+BQAB+BC\phantom{\rule{0ex}{0ex}}i.e.\left(AQ+CQ\right)+2BQAB+BC\phantom{\rule{0ex}{0ex}}i.e.AC+2BQAB+BC$

Regards

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