Answer 1

We have,

In ΔBAC, by Pythagoras theorem

BC2 = AB2 + AC2

⇒ BC2 = c2 + b2

⇒ BC = `sqrt(c^2+b^2)`             ........(i)

In ΔABD and ΔCBA

∠B = ∠B                             [Common]

∠ADB = ∠BAC                     [Each 90°]

Then, ΔABD ~ ΔCBA           [By AA similarity]

`therefore"AB"/"CB"="AD"/"CA"`           [Corresponding parts of similar Δ are proportional]

`rArrc/sqrt(c^2+b^2)="AD"/b`

`rArr"AD"="bc"/sqrt(c^2+b^2)`