Question

11) In the figure AB = BC and AD is
perpendicular to CD. Prove that :
AC^2 = 2.BC.DC. ​

Answer 1

Answer:

Step-by-step explanation:

Using Pythagoras Theorem in Δ ADB

AB² = AD² + DB²

AD² = AB² - DB² ---------eq(I)

Using Pythagoras Theorem in Δ ADC

AC² = AD² + DC²

AC² = (AB² - DB²)  + DC²------------[Putting the value of AD² from eq(I)]

AC² = (AB² - DB²)  + (DB+BC)²----------------[From the figure]

AC² = (AB² - DB²)  + DB² + BC² + 2BD.BC

AC² = Bc² + BC² + 2BD.BC -----[IF AB=BC then AB² = BC²]

AC² = 2BC (BC+BD)

AC² = 2BCDC  [Hence proved]