Question

Two right triangles ABC &DBC are drawn on the same side of BC. If AC & DB intersects at P, prove that AC*PC=BP*PD


class 10 triangle please help me​

Answer 1

Step-by-step explanation:

ΔABC, ΔDBC are right angle triangles at A and D respectively.  We use Pythagoras theorem in all the right angle Δs.  Also, ΔDPC is a right angle Δ at D.

CD² = CP² - DP² 

       = (CA + AP)² - (DB + BP)²

       = CA² + AP² + 2 CA AP - DB² - BP² - 2 DB BP

=> CD² + DB² = CA² - (BP² - AP²) + 2 CA * AP - 2 DB BP

=> CB² = CA² - AB² + 2 CA AP - 2 DB BP

=> 2 AB² = 2 CA * AP - 2 DB * BP              as CB² = CA² + AB²

=> AB² = (PC - AP) AP - (DP - BP) BP

            = AP* PC - AP² - DP * BP + BP²

            = AP  PC - DP * BP + AB²

=>  AP * PC = DP * BP

Answer 2

Answer:

here your answer

Step-by-step explanation:

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