
In $\Delta ABC,AD$ is the perpendicular bisector of BC. Show that $\Delta ABC$ is an isosceles triangle in which $AB=AC.$
Answer
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Hint:$AD\bot BC$ . Hence point D divides BC into 2 equal halves and $\angle ADB={{90}^{\circ }}$ . Consider $\Delta ADB\text{ and }\Delta ADC$ , prove their similarity by SAS congruence criteria. Take CPCT and prove $AB=AC.$
Complete step-by-step answer:
We have been given a $\Delta ABC$ and it is said that AD is perpendicular bisector of BC. Thus we can say that $AD\bot BC$ .
As $AD\bot BC$ from the figure we can say that $\angle ADB=\angle ADC={{90}^{\circ }}$ .
We have been given in the figure that $\angle ADB={{90}^{\circ }}$ then AD bisects line BC, as it is a perpendicular bisector. Thus we can say that line AD divides BC into two equal half. This means that BD is equal to DC.
i.e. $BD=DC$ .
Now , we have been asked to prove that $\Delta ABC$ is an isosceles triangle. For a triangle to be isosceles, two sides should be equal. i.e. sides AB and AC should be equal.
Now from the figure, let us consider $\Delta ABD$ and $\Delta ACD$ .
$AD=AD$ , this side is common to both the triangles.
$\angle ADB=\angle ADC$ , we said earlier that line $AD\bot BC$.
$BD=CD$ , we said that AD bisects BC , as it’s a perpendicular of bisector.
As two sides and one angle of $\Delta ABD$ is equal to two sides and one angle of $\Delta ACD$ , we can say that by SAS congruence rule both triangles are equal.
i.e. $\Delta ADC\cong \Delta ADB$ .
Thus by CPCT (corresponding parts of the congruent triangle) if two triangles are congruent to each other then the corresponding angles and the sides of the triangles are also congruent to each other.
Thus by CPCT rule we can say that,
$AB=AC$ .
Hence two sides of $\Delta ABC$ is equal. Therefore $\Delta ABC$ is an isosceles triangle.
$\therefore $ We proved that $\Delta ABC$ is an isosceles triangle.
Note: If two sides of a triangle are congruent then the angle opposite to those sides are congruent. If angles opposite to those sides are congruent then two sides of a triangle are congruent.
Complete step-by-step answer:
We have been given a $\Delta ABC$ and it is said that AD is perpendicular bisector of BC. Thus we can say that $AD\bot BC$ .
As $AD\bot BC$ from the figure we can say that $\angle ADB=\angle ADC={{90}^{\circ }}$ .
We have been given in the figure that $\angle ADB={{90}^{\circ }}$ then AD bisects line BC, as it is a perpendicular bisector. Thus we can say that line AD divides BC into two equal half. This means that BD is equal to DC.
i.e. $BD=DC$ .
Now , we have been asked to prove that $\Delta ABC$ is an isosceles triangle. For a triangle to be isosceles, two sides should be equal. i.e. sides AB and AC should be equal.
Now from the figure, let us consider $\Delta ABD$ and $\Delta ACD$ .
$AD=AD$ , this side is common to both the triangles.
$\angle ADB=\angle ADC$ , we said earlier that line $AD\bot BC$.
$BD=CD$ , we said that AD bisects BC , as it’s a perpendicular of bisector.
As two sides and one angle of $\Delta ABD$ is equal to two sides and one angle of $\Delta ACD$ , we can say that by SAS congruence rule both triangles are equal.
i.e. $\Delta ADC\cong \Delta ADB$ .
Thus by CPCT (corresponding parts of the congruent triangle) if two triangles are congruent to each other then the corresponding angles and the sides of the triangles are also congruent to each other.
Thus by CPCT rule we can say that,
$AB=AC$ .
Hence two sides of $\Delta ABC$ is equal. Therefore $\Delta ABC$ is an isosceles triangle.
$\therefore $ We proved that $\Delta ABC$ is an isosceles triangle.
Note: If two sides of a triangle are congruent then the angle opposite to those sides are congruent. If angles opposite to those sides are congruent then two sides of a triangle are congruent.
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