Question

The Side BC of Triangle ABC is Produced, Such that D is on ray BC.The bisector of Angle A meets BC in L.Prove that AngleABC+AngleACD=2AngleALC

Answer 1

the solution is above

Answer 2

Answer:

<ABL + <ACD = 2<ALC

Step-by-step explanation:

In ∆ABC , BC extended to D.

i) In ∆ALC ,

Exterior angle at C = sum of interior opposite angles

z = x+p ----(1)

ii) In ∆ABL ,

Exterior angle at L = x+y

=> p = x+y

=> y = p-x ---(2)

Now ,

Add equations (1) and (2), we get

y + z = p - x + x + p

=> y + z = 2p

=> <ABL + <ACD = 2<ALC

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