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dem82 [27]
3 years ago
9

title=" \underline{ \underline{ \text{Question}}} : " alt=" \underline{ \underline{ \text{Question}}} : " align="absmiddle" class="latex-formula">
In the adjoining figure , ABC is an isosceles triangle. BO and CO are the bisectors of \angle ABC and \angle ACB respectively. Prove that BOC is an isosceles triangle.

~Thanks in advance !

Mathematics
2 answers:
Marina CMI [18]3 years ago
8 0
<h3><u>GiveN</u><u> </u><u>:</u><u>-</u></h3>
  • In the given figure , ABC is an isosceles triangle.
  • BO and CO are the bisectors of \angle ABC and \angle ACB respectively.

<h3><u>To</u><u> </u><u>ProvE</u><u> </u><u>:</u><u>-</u></h3>

  • ∆ BOC is isosceles.

<h3><u>ProoF</u><u> </u><u>:</u><u>-</u></h3>

Here it's given that ∆ABC is a isosceles triangle. We know that in a isosceles triangle opposite angles are equal. And the angles opposite to equal sides are also equal.

Hence here ,

  • AB = AC
  • ∠ ABC = ∠ACB .

<u>Figure</u><u> </u><u>:</u><u>-</u>

\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\put(0,0){\line(1,0){4}}\put(4,0){\line(-1,2){2}}\put(0,0.001){\line(1,2){2}}\put(0,0.01){\line(1,1){2}}\put(3.99,0){\line(-1,1){2}}\put(0,-0.3){$\bf B $}\put(4,-0.3){$\bf C$}\put(2,4.2){$\bf A$}\put(1.8,2.2){$\bf o$}\end{picture}

If LaTeX doesn't work on app kindly see the attachment .

Here we will use a theorem which is ,

\large\underline{\underline{\textsf{\textbf{\red{\blue{$\leadsto$} Theorem :- }}}}}

  • Sides opposite to equal angles are equal.

Hence here,

\tt:\implies \angle ABC =\angle ACB \\\\\tt:\implies \dfrac{1}{2}\times \angle ABC =  \dfrac{1}{2}\times \angle ACB \\\\\tt:\implies \boxed{\bf \blue{\angle OBC = \angle OCB} }

Now in ∆OBC , we can see that ∠OBC = ∠OCB. (just proved ) . Hence we can say that sides OB and OC are equal. Since the sides opposite to equal angles are equal. Therefore in ∆OBC , two sides are equal.

Therefore ∆ OBC is an isosceles triangle.

<h3><u>Hence </u><u>Proved</u><u> </u><u>!</u><u> </u></h3>

Lubov Fominskaja [6]3 years ago
7 0

Answer:

See Below.

Step-by-step explanation:

We are given the isosceles triangle ΔABC. By the definition of isosceles triangles, this means that ∠ABC = ∠ACB.

Segments BO and CO bisects ∠ABC and ∠ACB.

And we want to prove that ΔBOC is an isosceles triangle.

Since BO and CO are the angle bisectors of ∠ABC and ∠ACB, respectively, it means that ∠ABO = ∠CBO and ∠ACO = ∠BCO.

And since ∠ABC = ∠ACB, this implies that:

∠ABO = ∠CBO =∠ACO = ∠BCO.

This is shown in the figure as each angle having only one tick mark, meaning that they are congruent.

So, we know that:

\angle ABC=\angle ACB

∠ABC is the sum of the angles ∠ABO and ∠CBO. Likewise, ∠ACB is the sum of the angles ∠ACO and ∠BCO. Hence:

\angle ABO+\angle CBO =\angle ACO+\angle BCO

Since ∠ABO =∠ACO, by substitution:

\angle ABO+\angle CBO =\angle ABO+\angle BCO

Subtracting ∠ABO from both sides produces:

\angle CBO=\angle BCO

So, we've proven that the two angles are congruent, thereby proving that ΔBOC is indeed an isosceles triangle.

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Answer:

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Step one:

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Answer:

<u>a) x = 3</u>

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Step-by-step explanation:

a) 2x = 6

Despejamos x dividiendo por 2 a amabos lados de la eacuacion.

(2/2)x = 6/2

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b) 10 + z = 20

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Si remplazamos z en la ecuación original:

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c) p + 9 = 11

Despejamos p restando 9 en amabos lados de la eacuacion.

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Si remplazamos p en la ecuación original:

2 + 9 = 11

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Despejamos x restando 8 en amabos lados de la eacuacion y luego divideindo por 3 en ambos lados de la ecuación.

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Si remplazamos x en la ecuación original:

3(7) + 8 = 29

21 + 8 = 29

29 = 29

Queda demostrado

e) 2u + 8 = 10

Despejamos u restando 8 en amabos lados de la eacuacion y luego divideindo por 2 en ambos lados de la ecuación.

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(2/2)x = 2/2

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Si remplazamos x en la ecuación original:

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Espero te haya sido de ayuda!

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