Question

Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side BC If AC and BD intersects at P prove that AP*PC=BP*PD

Answer 1

See diagram.
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:  by the given figure,

in triangle BPA and triangle CPD

angle BPA = angle CPD    ( V.O.A)

angle BAC = angle CDB    ( 90 degree each )

therefore,

triangle PBA similar triangle CPD

AP/DP = PB/PC     (by bpt)

hence AP*PC=BP*PD

Step-by-step explanation: