From a point O within a tringle ABC, I have drawn the perpendiculars OX, OY and OZ on BC, CA and AB respectively. Let us prove that, AZ² + BX² + CY² = AY² + CX² + BZ²

By applying Pythagoras theorem in ΔAOZ, we get,

AO² = AZ² + ZO²

⇒ AZ² = AO² - ZO² ……………… (1)

By applying Pythagoras theorem in ΔBOX, we get,

BO² = BX² + XO²

⇒ BX² = BO² - XO² ………………… (2)

By applying Pythagoras theorem in ΔCOY, we get,

CO² = CY² + YO²

⇒ CY² = CO² - YO² ………………… (3)

By adding eqn. (1), (2) and (3), we have,

AZ² + BX² + CY² = AO² - ZO² + BO² - XO² + CO² - YO²

⇒ AZ² + BX² + CY² = AO² + BO² + CO² - XO² - YO² - ZO² …… (i)

By applying Pythagoras theorem in ΔAOY, we get,

AO² = AY² + YO²

⇒ AY² = AO² - YO² ………… (4)

By applying Pythagoras theorem in ΔBOZ, we get,

BO² = BZ² + ZO²

⇒ BZ² = BO² - ZO² …………… (5)

By applying Pythagoras theorem in ΔCOX, we get,

CO² = CX² + XO²

⇒ CX² = CO² - XO² …………… (6)

By adding eqn. (4), (5) and (6), we have,

AY² + BZ² + CX² = AO² - YO² + BO² - ZO² + CO² - XO²

⇒ AY² + CX² + BZ² = AO² + BO² + CO² - XO² - YO² - ZO² …… (ii)

By comparing eqn. (i) and (ii) we get,

AZ² + BX² + CY² = AY² + CX² + BZ²

Hence proved.