P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that ${A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$

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Solution

In right- angled triangles ACQ and PCB, we have

${A Q}^{2}$ = ${A C}^{2}$ + ${C Q}^{2}$ and ${P B}^{2}$ = ${P C}^{2}$ + ${C B}^{2}$

⇒ ${A Q}^{2}$ + ${B P}^{2}$ = ( ${A C}^{2}$ + ${C Q}^{2}$ ) + ( ${P C}^{2}$ + ${C B}^{2}$ )

⇒ ${A Q}^{2}$ + ${B P}^{2}$ = ( ${A C}^{2}$ + ${B C}^{2}$ ) + ( ${P C}^{2}$ + ${Q C}^{2}$ )

⇒ ${A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$

${A C}^{2}$ + ${B C}^{2}$ - ${A B}^{2}$ and ${P C}^{2}$ + ${Q C}^{2}$ - ${P Q}^{2}$ [By Pythagoras theorem]

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P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that AQ2 + BP2 = AB2 + PQ2

In right- angled triangles ACQ and PCB, we have AQ2 = AC2 + CQ2 and PB2 = PC2 + CB2 ⇒AQ2 + BP2 = (AC2 + CQ2) + (PC2 + CB2) ⇒AQ2 + BP2 = (AC2 + BC2) + (PC2 + QC2) ⇒AQ2 + BP2 = AB2 + PQ2 AC2 + BC2 - AB2 and PC2 + QC2 - PQ2 [By Pythagoras theorem]

P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that ${A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$