Question

Q 40 only from this page.Other questions are easy but this one question i am unable to understand.If somebody actually did the whole page then also i had be thankful.

Answer 1

GIVEN:-

• $\rm{\angle{B} = 2\angle{C}}$

• AD Bisect $\rm{\angle{BAC}}$ and AB = CD

• BE is the Bisector of $\rm{\angle{B}}$

TO FIND :-

• The measure of $\rm{\angle{BAC}}$

CONSTRUCTION :-

• Join D to E,

In ∆ABC ,we have $\rm{\angle{B} = 2\angle{C}}$ or $\rm{\angle{B} = 2y}$, where $\rm{\angle{C} = y}$.

AD is the Bisector of $\rm{\angle{BAC}}$ So, Let $\rm{\angle{BAD} =\angle{CAD} = x}$

Now, In ∆ ABE and ∆ DEC

$\implies\rm{AB = CD}$ (Given).

$\implies\rm{ BE = CE }$ (Given).

$\implies\rm{\angle{ABP} =\angle{DCE} = x}$(Proved Above).

Therefore, ∆ ABE ≈ ∆ DEC By S-A-S congurence Criteria.

Through this, AE = DE, $\rm{\angle{BAE} =\angle{CDE}}$. (C.P.C.T).

So,

$\implies\rm{\angle{CDP} = 2x}$ (A = 2x).

$\implies\rm{\angle{ADP} =\angle{DAP} = x}$

Now, In ∆ ABD, we have.

$\implies\rm{\angle{ADC} = \angle{ABD} +\angle{BAD}}$

$\implies\rm{ x + 2x = 2y + x }$

$\implies\rm{ x = y }$

In ∆ ABC , we have

$\implies\rm{\angle{A} + \angle{B} + \angle{C} = 180^{\circ}}$

$\implies\rm{ 2x + 2y + y = 180^{\circ}}$

$\implies\rm{ 2y + 2y + y = 180^{\circ}}$

$\implies\rm{ 5y = 180^{\circ}}$

$\implies\rm{ y = \dfrac{180}{5}}$

$\implies\rm{ y = 36}$

Hence, $\rm{\angle{BAC} = 2x = 72^{\circ}}$.