#### 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

AB^{2} = AC^{2} + BC^{2}

⇒ AB^{2} = (5)^{2} + (12)^{2}

⇒ AB^{2} = 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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#### APPEARS IN

Solution In Below Figure, triangle Abc is Right Angled at C and De ⊥ Ab. Prove that δAbc ~ δAde and Hence Find the Lengths of Ae and De. Concept: Triangles Examples and Solutions.