Question

In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE

Answer 1

In ∆ABC and ∆ADE

∠ACB = ∠AED = 90°

∠A = ∠A  ...(common)

∴ ∆ABC ~ ∆ADE  ...(By AA similarity)

`"BC"/"DE" = "AB"/"AD" = "AC"/"AE"`

`12/"DE" = 13/3 = 5/"AE"`

In ∆ABC, AB² = BC² + AC²

= 122 + 52

= 144 + 25

= 169

AB = `sqrt(169)` = 13

Consider, `13/3 = 5/"AE"`

∴ AE = `(5 xx 3)/13 = 15/13`

AE = `15/13` and DE = `36/13`

Consider, `12/"DE" = 13/3`

DE = `(12 xx 3)/13 = 36/13`