In ΔABC I have drawn two perpendiculars from two vertices B and C on AC and AB (AC>AB) which are intersected each other at the point P. Let us prove that, AC2 + BP2 = AB2 + CP2
Given: In ΔABC, AC>AB
CE ⊥ AB
BD ⊥ AC

To prove: AC2 + BP2 = AB2 + CP2
Proof:
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