In Fig. 4.145 (a) ΔABC is right angled at C and DE⊥AB. Prove that ΔABC~ ΔADE and hence find the lengths of AE and DE.
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In ΔACB
ABsq.= ACsq.+ BCsq.
ABsq.= 25+ 144 = 169
AB= √169 = 13cm
In ΔAED and ΔACB
∠A is common in both
∠AED = ∠ACB (90 degree)
ΔAED~ΔACB
Therefore, AE/AC = DE/CB = AD/AB (corresponding parts of similar Δ are proportional)
AE/5 =3/13 and DE/12 =3/13
AE = 15/13 cm and DE = 36/13cm
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