Question

2. Two right triangles ABC and DBC are drawn on the
same hypotenuse BC and on the same side of BC.
If AC and BD intersect at P, prove that AP PC
= BP ~ DP.
[3]​

Answer 1

$\huge{\underline{\underline{\mathfrak{QUESTION:}}}}$

Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × DP.

$\huge{\underline{\underline{\mathfrak{\green{ANSWER:}}}}}$

In ∆BPA & ∆CPD,

$\implies$ $\angle$BAP = $\angle$CDP = 90°

$\implies$ $\angle$APB = $\angle$DPC (Vertically opposite angles)

$\therefore$ ∆BPA ~ ∆CPD (By AA criterion)

$\implies$ AD/DP = PB/PC

$\implies$ AD × PC = BP × DP

Hence Proved!

_________________________

Answer 2

Answer:

Hey mate here is ur ans

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