In triangle abc, ad is a median and am is perpendicular to bc show that ab^2+ac^2=2(ad^2+bd^2)
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We have,
ab² =am²+bm² ,ac²= am²+cm²
ab² +ac²=2am²+bm²+cm²
Also,
bm²+cm=(bd-dm)²+(cd+dm)²=2bd²+2dm²(using bd=cd)
Now,
2am²+bm²+cm²=2bd²+2(dm²+am²)=2(bd²+ad²)
ab² =am²+bm² ,ac²= am²+cm²
ab² +ac²=2am²+bm²+cm²
Also,
bm²+cm=(bd-dm)²+(cd+dm)²=2bd²+2dm²(using bd=cd)
Now,
2am²+bm²+cm²=2bd²+2(dm²+am²)=2(bd²+ad²)
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