Question

The bisectors of the angles A, B of ∆ABC intersect in I, the bisectors of the corresponding exterior
angles intersect in E. Prove that: AIBE is cyclic.
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Answer 1

ALBE is cyclic if  The bisectors of the angles A, B of ∆ABC intersect in L, the bisectors of the corresponding exterior angles intersect in E

Step-by-step explanation:

The bisectors of the angles A, B of ∆ABC intersect in L

=> ∠BAL = ∠A/2    & ∠ABL  = ∠B/2

the bisectors of the corresponding exterior angles intersect in E

Exterior of ∠A = ∠B + ∠C    & Exterior of ∠B = ∠A + ∠C

=> ∠BAE = (∠B + ∠C) /2    & ∠ABE  =  (∠A + ∠C) /2

∠EAL = ∠BAL + ∠BAE = ∠A/2  +  (∠B + ∠C) /2   = (∠A + ∠B + ∠C) /2

∠EBL = ∠ABL + ∠ABE = ∠B/2  +  (∠A + ∠C) /2   = (∠A + ∠B + ∠C) /2

∠EAL + ∠EBL  = (∠A + ∠B + ∠C) /2 +  (∠A + ∠B + ∠C) /2

=> ∠EAL + ∠EBL  = ∠A + ∠B + ∠C

=> ∠EAL + ∠EBL  = 180°

as sum of opposite angle of ALBE is 180° Hence ALBE is cyclic

QED

Proved

Learn more:

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Answer 2

Solution:

proof: Take point P&Q

m angle CAB +m angle BAP ••••• linear pairs axiom

therefore half m angle CAB +m angle BAP =half ×180°•••••multiplying by half

for further ans the attachment is provided

hope this helps you mate