Question

please solve this question If ABC Is a right triangle such that AB=AC and bisector of angle c intersect side ab at d then prove ac + ad = bc

Answer 1

Draw a perpendicular from D onto BC to meet it at E.
∠EDC = ∠ADC   as  ∠E = 90° = ∠A  and   ∠ACD = ∠ECD as CD is a bisector of ∠C.

ΔADC and ΔEDC are congruent, as CD is a common side.
      =>  AC = CE         and      AD = DE

In ΔBDE, ∠BDE = ∠DBE 
       
=>  BE = AD

AC + AD = CE + EB
               = BC

Answer 2

 Sol:  Let AB = AC = a and AD = b In a right angled triangle ABC , BC2 = AB2 + AC2 BC2 = a2 + a2 BC = a√2 Given AD = b, we get DB = AB – AD or DB = a – b We have to prove that AC + AD = BC or (a + b) = a√2. By the angle bisector theorem, we get AD/ DB = AC / BC b/(a - b) = a/ a√2 b/(a - b) = 1/√2 b = (a – b)/ √2 b√2 = a – b b(1 + √2) = a b = a/ (1 + √2) Rationalizing the denominator with (1 - √2)   b = a(1 - √2) / (1 + √2) × (1 - √2) b = a(1 - √2)/ (-1) b = a(√2 - 1) b = a√2 – a b + a = a√2 or  AD + AC = BC [we know that AC = a, AD = b and BC = a√2] Hence it is proved. Plz if u like mark it