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Question
In below Figure, ΔABC is right angled at C and DE ⊥ AB. Prove that ΔABC ~ ΔADE and Hence find the lengths of AE and DE.
Solution
In ΔACB, by Pythagoras theorem
AB2 = AC2 + BC2
⇒ AB2 = (5)2 + (12)2
⇒ AB2 = 25 + 144 = 169
⇒ AB = `sqrt169` = 13 cm
In ΔAED and ΔACB
∠A = ∠A [Common]
∠AED = ∠ACB [Each 90°]
Then, ΔAED ~ ΔACB [By AA similarity]
`therefore"AE"/"AC"="DE"/"CB"="AD"/"AB"` [Corresponding parts of similar Δ are proportional]
`rArr"AE"/5="DE"/12=3/13`
`rArr"AE"/5=3/13` and `"DE"/12=3/13`
`"AE"=15/13` cm and `"DE"=36/13` cm
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