Question Number 186838 by mnjuly1970 last updated on 11/Feb/23
Answered by mr W last updated on 11/Feb/23
Commented by mr W last updated on 11/Feb/23
$${make}\:{PD}={PB},\:{BP}'={BP} \\ $$$$\Rightarrow{P}'{C}={PA} \\ $$$$\Delta{BDC}\equiv\Delta{BP}'{C} \\ $$$$\Rightarrow{DC}={P}'{C}={PA} \\ $$$${PC}={PD}+{DC}={PB}+{PA}\:\checkmark \\ $$
Answered by mr W last updated on 11/Feb/23
Commented by mr W last updated on 11/Feb/23
$${s}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\:{ac}\:\:\:…\left({i}\right) \\ $$$${s}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{bc}\:\:\:…\left({ii}\right) \\ $$$${s}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\:\:\:…\left({iii}\right) \\ $$$$\left({iii}\right)×\mathrm{2}−\left({i}\right)−\left({ii}\right): \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{ab}+{ac}+{bc}=\mathrm{0} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +\left({a}+{b}\right){c}−\mathrm{2}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}+{b}+\mathrm{2}{c}\right)\left({a}+{b}−{c}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}+{b}−{c}=\mathrm{0}\: \\ $$$$\Rightarrow{c}={a}+{b}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 11/Feb/23
$${very}\:{nice}\:{solution}\:{sir}\:{W}… \\ $$
Answered by som(math1967) last updated on 11/Feb/23
![cut PX such PX=PA now PX=PA ∠APC=∠ABC=60 △PAX equilateral ∴ AP=PX=AX ∠BAP+∠BAX=∠BAX+∠XAC=60 ∠BAP=∠XAC △APB≅△CXA [AB=AC,∠BAP=∠XAC ,PA=AX] ∴CX=PB ∴CP=CX+XP =PA+PB](https://www.tinkutara.com/question/Q186845.png)
$${cut}\:{PX}\:{such}\:{PX}={PA} \\ $$$${now}\:{PX}={PA} \\ $$$$\angle{APC}=\angle{ABC}=\mathrm{60} \\ $$$$\bigtriangleup{PAX}\:{equilateral} \\ $$$$\therefore\:{AP}={PX}={AX} \\ $$$$\angle{BAP}+\angle{BAX}=\angle{BAX}+\angle{XAC}=\mathrm{60} \\ $$$$\angle{BAP}=\angle{XAC} \\ $$$$\bigtriangleup{APB}\cong\bigtriangleup{CXA} \\ $$$$\left[{AB}={AC},\angle{BAP}=\angle{XAC}\:,{PA}={AX}\right] \\ $$$$\therefore{CX}={PB} \\ $$$$\therefore{CP}={CX}+{XP} \\ $$$$={PA}+{PB} \\ $$
Commented by som(math1967) last updated on 11/Feb/23
