Question Number 225840 by hardmath last updated on 14/Nov/25
Commented by hardmath last updated on 14/Nov/25
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{MA}^{\mathrm{2}} +\mathrm{MB}^{\mathrm{2}} +\mathrm{MC}^{\mathrm{2}} +\mathrm{MD}^{\mathrm{2}} \:=\:\mathrm{4R}^{\mathrm{2}} \\ $$
Answered by A5T last updated on 14/Nov/25
![[ABC]=((MC×AB)/2)=((AC×AB×BC)/(4R)) ⇒MC=(((√(MC^2 +AM^2 ))×(√(MC^2 +MB^2 )))/(2R)) MC×MD=AM×MB ⇒4R^2 =((MC^4 +MB^2 MC^2 +AM^2 MC^2 +AM^2 MB^2 )/(MC^2 )) ⇒4R^2 =MC^2 +MB^2 +AM^2 +MD^2](https://www.tinkutara.com/question/Q225850.png)
$$\left[\mathrm{ABC}\right]=\frac{\mathrm{MC}×\mathrm{AB}}{\mathrm{2}}=\frac{\mathrm{AC}×\mathrm{AB}×\mathrm{BC}}{\mathrm{4R}} \\ $$$$\Rightarrow\mathrm{MC}=\frac{\sqrt{\mathrm{MC}^{\mathrm{2}} +\mathrm{AM}^{\mathrm{2}} }×\sqrt{\mathrm{MC}^{\mathrm{2}} +\mathrm{MB}^{\mathrm{2}} }}{\mathrm{2R}} \\ $$$$\mathrm{MC}×\mathrm{MD}=\mathrm{AM}×\mathrm{MB} \\ $$$$\Rightarrow\mathrm{4R}^{\mathrm{2}} =\frac{\mathrm{MC}^{\mathrm{4}} +\mathrm{MB}^{\mathrm{2}} \mathrm{MC}^{\mathrm{2}} +\mathrm{AM}^{\mathrm{2}} \mathrm{MC}^{\mathrm{2}} +\mathrm{AM}^{\mathrm{2}} \mathrm{MB}^{\mathrm{2}} }{\mathrm{MC}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{4R}^{\mathrm{2}} =\mathrm{MC}^{\mathrm{2}} +\mathrm{MB}^{\mathrm{2}} +\mathrm{AM}^{\mathrm{2}} +\mathrm{MD}^{\mathrm{2}} \\ $$