Question

The Side BC of the triangle A B C is reduced to D, the bisector of angle A meets BC is L, Prove That ∠ABC + ∠ACD = 2∠ALC

Answer 1

Corrected Question:

• The side BC of the ∆ABC is produced to D. The bisector of ∠A meets BC in L. Prove that ∠ABC + ∠ACD = 2∠ALC.

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Given: The side BC of the ∆ABC is produced to D. The bisector of ∠A meets BC in L.

Need to Prove: ∠ABC + ∠ACD = 2∠ALC.

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• We know, if a side of a triangle is produced, then the exterior angle so formed is equal to the sum of two interior opposite angles. So,

$\\\; \; :\implies\quad\sf{\angle ACD=\angle ABC+\angle BAC\qquad\qquad\left\lgroup\begin{matrix}\sf{{eq}^{n}\: (1)}\end{matrix}\right\rgroup\]}$

Its is given in the question that AL is bisector of ∠BAC. Then,

$\\\; \; :\implies\quad\sf{\angle BAC=\angle BAL}$

Substituting value of ∠BAC in eqⁿ ( 1 ).

$\\\; \; :\implies\quad\sf{\angle ACD=\angle ABC+2\angle BAL}$

Adding ∠ABC on both side.

$\\\; \; :\implies\quad\sf{\angle ABC+\angle ACD=\angle ABC +\angle ABC +2\angle BAL}$

Performing addition.

$\\\; \; :\implies\quad\sf{\angle ABC+\angle ACD=2\angle ABC+2\angle BAL}$

Taking 2 common in RHS.

$\\\; \; :\implies\quad\sf{\angle ABC+\angle ACD=2(\angle ABC+\angle BAL)}$

Applying exterior angle theorem in ∆ABL.

$\\\; \; :\implies\quad\underline{\boxed{\bold{\angle ABC+\angle ACD=2\angle ALC}}}$

Hence, Proved!