Question

plzz solve question no 6,1st one​

Answer 1

Here, CD = BD - BC

And BD = BC + CD

Both triangle ABD and ACD are right triangles.

So in ΔACD, AD² = AC² - CD²   →   (1)

In ΔABD,

AB² = BD² + AD²

AB² = BD² + (AC² - CD²)                                  [According to (1)]

AB² = BD² + AC² - CD²

AB² = BD² + AC² - (BD - BC)²                          [CD = BD - BC]

AB² = BD² + AC² - (BD² - 2BC · BD + BC²)

AB² = BD² + AC² - BD² + 2BC · BD - BC²

AB² = AC² + 2BC · BD - BC²

AB² = AC² - BC² + 2BC · BD

AB² = AC² - BC² + 2BC(BC + CD)                    [BD = BC + CD]

AB² = AC² - BC² + 2BC² + 2BC · CD

AB² = AC² + BC² + 2BC · CD

AB² = BC² + AC² + 2BC · CD

Hence proved!

Hope this helps. ^_^

Thank you. :-))