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`        If x(b+c)=y(c+a)=z(a+b), then prove that a÷(y+z-x)=b÷(z+x-y)=c÷(x+y-z)`
one year ago

```							Let us see:x/(a+b-c) = y/(b+c-a) = z(c+a-b) = k=> x = ka+kb-kc —————————→ 1and y = kb+kc-ka —————————→ 2and z = kc+ka-kb —————————-> 3add 1 and 3 above to get z+x = 2ka => (z+x)/a = 2k ————————-> 4add 1 and 2 above to get x+y = 2kb => (x+y)/b = 2k ————————→ 5add 2 and 3 above to get y+z = 2kc => (y+z)/c = 2k —————————→ 6from 4, 5 and 6 above, we have (z+x)/a = (x+y)/b = (y+z)/c =====> PROVED
```
one year ago
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