
Prove that the bisectors of the base angles of an isosceles triangle are equal.
Answer
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Hint: Bisector of an angle is the line making the half of the angle and we are asked to find the bisectors of the base angles of an isosceles triangle. We can prove this by proving the two triangles corresponding to the base angles of the isosceles triangle as congruent.
Complete step-by-step answer:
We are given an isosceles triangle, with two sides equal to each other and we have to prove that the bisectors of the angles corresponding to the two equal sides are equal to each other.
Suppose we have a $\Delta ABC$ with side $AB = AC$ as shown
Now by the property of isosceles triangle which states that “ Angles opposite to equal sides of the triangle are equal” we can say that,
$\angle B\, = \,\angle C\,\,\,\,\,\,\,\,....\left( i \right)$
Now both the angles are equal so their halves will also be equal as,
$ \Rightarrow \,\,\dfrac{1}{2}\angle B\, = \dfrac{1}{2}\angle C$
$ \Rightarrow \,\,\angle CBE\, = \,\angle BCF\,\,\,\,\,\,....\left( {ii} \right)$
Now we have to prove the two triangles as congruent,
So we have two triangles as $\Delta BCF$ and $\Delta CBE$
Here in both the triangles we have,
$\angle B\, = \,\angle C$ [from $\left( i \right)$ ]
$\angle BCF = \angle CBE$ [from $\left( {ii} \right)$ ]
Side $BC = BC$ [Common in both]
Then by Angle Angle Side (AAS) property of triangle, we can say that
$\Delta BCF \cong \,\,\Delta CBE$ [both these triangles are congruent to each other]
Then by corresponding parts of congruent triangle that is, by c.p.c.t we can say that
$ \Rightarrow \,\,BE = CF$
Thus we have proved that both the bisectors $BE$ and $CF$ of an isosceles triangle $\Delta ABC$ are equal.
Note: There are various properties of triangles which are used to prove the two triangles as congruent to each other. These properties include SSS(side, side, side), SAS(side, angle, side), ASA(angle, side, angle), AAS(angle, angle, side) etc. These all are used according to the conditions mentioned in the triangle.
Make sure to use the right property of congruence to prove the two triangles as congruent. Firstly carefully examine the triangles to be proved as congruent, then apply the property accordingly which suits the conditions of the triangle.
Complete step-by-step answer:
We are given an isosceles triangle, with two sides equal to each other and we have to prove that the bisectors of the angles corresponding to the two equal sides are equal to each other.
Suppose we have a $\Delta ABC$ with side $AB = AC$ as shown
Now by the property of isosceles triangle which states that “ Angles opposite to equal sides of the triangle are equal” we can say that,
$\angle B\, = \,\angle C\,\,\,\,\,\,\,\,....\left( i \right)$
Now both the angles are equal so their halves will also be equal as,
$ \Rightarrow \,\,\dfrac{1}{2}\angle B\, = \dfrac{1}{2}\angle C$
$ \Rightarrow \,\,\angle CBE\, = \,\angle BCF\,\,\,\,\,\,....\left( {ii} \right)$
Now we have to prove the two triangles as congruent,
So we have two triangles as $\Delta BCF$ and $\Delta CBE$
Here in both the triangles we have,
$\angle B\, = \,\angle C$ [from $\left( i \right)$ ]
$\angle BCF = \angle CBE$ [from $\left( {ii} \right)$ ]
Side $BC = BC$ [Common in both]
Then by Angle Angle Side (AAS) property of triangle, we can say that
$\Delta BCF \cong \,\,\Delta CBE$ [both these triangles are congruent to each other]
Then by corresponding parts of congruent triangle that is, by c.p.c.t we can say that
$ \Rightarrow \,\,BE = CF$
Thus we have proved that both the bisectors $BE$ and $CF$ of an isosceles triangle $\Delta ABC$ are equal.
Note: There are various properties of triangles which are used to prove the two triangles as congruent to each other. These properties include SSS(side, side, side), SAS(side, angle, side), ASA(angle, side, angle), AAS(angle, angle, side) etc. These all are used according to the conditions mentioned in the triangle.
Make sure to use the right property of congruence to prove the two triangles as congruent. Firstly carefully examine the triangles to be proved as congruent, then apply the property accordingly which suits the conditions of the triangle.
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