Question Number 221196 by mnjuly1970 last updated on 26/May/25
Answered by mr W last updated on 27/May/25
Commented by mr W last updated on 27/May/25
![AM=MB ⇒h_3 =((h_1 +h_2 )/2) ΔADN=ΔARD+ΔDRN=((DN×h_1 )/2)=((DC×h_1 )/4) ΔNBC=ΔPBC+ΔNPC=((NC×h_2 )/2)=((DC×h_2 )/4) ΔARD+ΔDRN+ΔPBC+ΔNPC=((DC×(h_1 +h_2 ))/4)=((DC×h_3 )/2) ΔDMC=ΔDRN+ΔNPC+[MPNR]=((DC×h_3 )/2) ΔARD+ΔDRN+ΔPBC+ΔNPC=ΔDRN+ΔNPC+[MPNR] ⇒ΔARD+ΔPBC=[MPNR]](https://www.tinkutara.com/question/Q221209.png)
$${AM}={MB}\:\Rightarrow{h}_{\mathrm{3}} =\frac{{h}_{\mathrm{1}} +{h}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Delta{ADN}=\Delta{ARD}+\Delta{DRN}=\frac{{DN}×{h}_{\mathrm{1}} }{\mathrm{2}}=\frac{{DC}×{h}_{\mathrm{1}} }{\mathrm{4}} \\ $$$$\Delta{NBC}=\Delta{PBC}+\Delta{NPC}=\frac{{NC}×{h}_{\mathrm{2}} }{\mathrm{2}}=\frac{{DC}×{h}_{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Delta{ARD}+\Delta{DRN}+\Delta{PBC}+\Delta{NPC}=\frac{{DC}×\left({h}_{\mathrm{1}} +{h}_{\mathrm{2}} \right)}{\mathrm{4}}=\frac{{DC}×{h}_{\mathrm{3}} }{\mathrm{2}} \\ $$$$\Delta{DMC}=\Delta{DRN}+\Delta{NPC}+\left[{MPNR}\right]=\frac{{DC}×{h}_{\mathrm{3}} }{\mathrm{2}} \\ $$$$\Delta{ARD}+\Delta{DRN}+\Delta{PBC}+\Delta{NPC}=\Delta{DRN}+\Delta{NPC}+\left[{MPNR}\right] \\ $$$$\Rightarrow\Delta{ARD}+\Delta{PBC}=\left[{MPNR}\right] \\ $$
Commented by mnjuly1970 last updated on 27/May/25
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