Question

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The question is in the attachment

Answer 1

Explanation:

a) Since CD is perpendicular to AB,

∠BDC = ∠CDA = 90°

Comparing ΔABC and ΔACD,

∠BCA = ∠CDA = 90°

∠CAB = ∠DAC (same angle)

since two angle are same in both triangles, the third angles will also be same

∠ABC = ∠ACD

∴ ΔABC and ΔACD are similar

Comparing ΔABC and ΔCBD,

∠BCA = ∠BDC = 90°

∠ABC = ∠CBD(same angle)

since two angle are same in both triangles, the third angles will also be same

∠CAB = ∠DCB

∴ ΔABC and ΔCBD are similar

b) AB = c, AC = a and BC = b

ΔABC and ΔACD are similar

`(AB)/(AC) =(AC)/(AD) =(BC)/(CD) \\\\(c)/(a) =(a)/(AD) =(b)/(CD) \\\\(c)/(a) =(a)/(AD)`

⇒ a² = c*AD - eq(1)

ΔABC and ΔCBD are similar

`(AB)/(CB) =(AC)/(CD) =(BC)/(BD) \\\\(c)/(b) =(a)/(CD) =(b)/(BD) \\\\(c)/(b) =(b)/(BD)`

⇒ b² = c*BD - eq(2)

eq(1) + eq(2):

(a² = c*AD ) + (b² = c*BD)

a² + b² = c*AD + c*BD

a² + b² = c*(AD + BD)

a² + b² = c*(c)

a² + b² = c²

Answer 2

I have completed it and attached in the explanation part.

Explanation: