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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

$$makePD=PB,{BP}^{\prime }=BP$$$$\Rightarrow P^{\prime }C=PA$$$$\mathrm{\Delta }BDC\equiv \mathrm{\Delta }{BP}^{\prime }C$$$$\Rightarrow DC=P^{\prime }C=PA$$$$PC=PD+DC=PB+PA✓$$

Answered by mr W last updated on 11/Feb/23

Commented by mr W last updated on 11/Feb/23

$$s^2=a^2+c^2-ac...\left(i\right)$$$$s^2=b^2+c^2-bc...\left(ii\right)$$$$s^2=a^2+b^2+ab...\left(iii\right)$$$$\left(iii\right)\times 2-\left(i\right)-\left(ii\right):$$$$a^2+b^2-2c^2+2ab+ac+bc=0$$$${(a+b)}^2+(a+b)c-2c^2=0$$$$(a+b+2c)(a+b-c)=0$$$$\Rightarrow a+b-c=0$$$$\Rightarrow c=a+b✓$$

Commented by mnjuly1970 last updated on 11/Feb/23

$$verynicesolutionsirW...$$

Answered by som(math1967) last updated on 11/Feb/23

$$cutPXsuchPX=PA$$$$nowPX=PA$$$$\mathrm{\angle }APC=\mathrm{\angle }ABC=60$$$$△PAXequilateral$$$$\therefore AP=PX=AX$$$$\mathrm{\angle }BAP+\mathrm{\angle }BAX=\mathrm{\angle }BAX+\mathrm{\angle }XAC=60$$$$\mathrm{\angle }BAP=\mathrm{\angle }XAC$$$$△APB\cong △CXA$$$$[AB=AC,\mathrm{\angle }BAP=\mathrm{\angle }XAC,PA=AX]$$$$\therefore CX=PB$$$$\therefore CP=CX+XP$$$$=PA+PB$$

Commented by som(math1967) last updated on 11/Feb/23

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