5 Marks Questions

Question 15 Marks

Bisectors of the angles $B$ and $C$ of an isosceles $\triangle\text{ABC}$ with $AB = AC$ intersect each other at $O.$ Show that external angle adjacent to $ \angle\text{ABC} $ is equal to $\angle\text{BOC.}$

Answer

Given Lines, $BO$ and $CO$ are angle bisectore isosceles such that $AB = AC$ which
$ \angle\text{ABC} $ and $\angle\text{ACB}$ respectively at $O.$
$\angle\text{DBA}=\angle\text{BOC}$ In $\triangle\text{ABC},$
$\text{AB}=\text{AC}$
$\angle\text{ACB}=\angle\text{ABC}$
$\Rightarrow \frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{ABC}$
$\Rightarrow \angle\text{OCB}=\angle\text{OBC}$

In $\triangle\text{OCB},$
$\angle\text{OCB}+\angle\text{OCB}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow\angle\text{OBC}+\angle\text{OBC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow 2\angle\text{OBC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow \angle\text{ABC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow 180^{\circ}-\angle\text{DBA}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow\angle\text{DBA}+\angle\text{BOC}=0$
$\Rightarrow \angle\text{DBA}=\angle\text{BOC}$

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Question 25 Marks

$M$ is a point on side $BC$ of a triangle $ABC$ such that $AM$ is the bisector of $\angle\text{BCA}.$ Is it true to say that perimeter of the triangle is greater than $2AM?$ Give reason for your answer$?$

Answer

In $\triangle\text{ABC,} M$ is point of side $BC$ such that $AM$ is the bisector $\text{AB}+\text{BM}>\text{AM}\ ....(\text{i})$ In $\triangle\text{ACM,}$ we have $\text{AC}+\text{CM}>\text{AM}\ ...(\text{ii})$

On adding eq.$(i)$ and $(ii),$
$\Rightarrow(\text{AB}+\text{BM}+\text{AC}+\text{CM})>2\text{AM}$
$\Rightarrow(\text{AB}+\text{BM}+\text{CM}+\text{AC})>2\text{AM}$
$\Rightarrow\text{AB}+\text{BC}+\text{AC}>2\text{AM}$

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