In ΔABC, If AP ⊥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB

1.	In ΔABC, If AP ⊥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB × CP.
Answer» Answer: In TRIANGLE abc,if AP is PERPENDICULAR to BC and ac^2=bc^2-ab^2 then prove that pa^2=pb×cp. 2.

In ΔABC, If AP ⊥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB × CP.

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In TRIANGLE abc,if AP is PERPENDICULAR to BC and ac^2=bc^2-ab^2 then prove that pa^2=pb×cp. 2.

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In ΔABC, If AP ⊥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB × CP.​

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In ΔABC, If AP ⊥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB × CP.