$ABC$ is a triangle. Locate a point in the interior of $\Delta ABC$ which is equidistant from all the vertices of $\Delta ABC$.
(N/A) Let us consider a $\Delta ABC$.
Draw line $l$ as the perpendicular bisector of side $AB$.
Draw line $m$ as the perpendicular bisector of side $BC$.
Let the two perpendicular bisectors $l$ and $m$ intersect at point $O$.
Since any point on the perpendicular bisector of a segment is equidistant from its endpoints,point $O$ lies on the perpendicular bisector of $AB$ (so $OA = OB$) and on the perpendicular bisector of $BC$ (so $OB = OC$).
Therefore,$OA = OB = OC$,which means point $O$ is the required point equidistant from all vertices $A, B,$ and $C$. This point $O$ is known as the circumcenter of the triangle.
Draw line $l$ as the perpendicular bisector of side $AB$.
Draw line $m$ as the perpendicular bisector of side $BC$.
Let the two perpendicular bisectors $l$ and $m$ intersect at point $O$.
Since any point on the perpendicular bisector of a segment is equidistant from its endpoints,point $O$ lies on the perpendicular bisector of $AB$ (so $OA = OB$) and on the perpendicular bisector of $BC$ (so $OB = OC$).
Therefore,$OA = OB = OC$,which means point $O$ is the required point equidistant from all vertices $A, B,$ and $C$. This point $O$ is known as the circumcenter of the triangle.
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Similar Questions
In the figure,$OA = OB$ and $OD = OC$. Show that
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.
Medium
View Solution$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ bisects $\angle A$.
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ bisects $\angle A$.
Medium
View SolutionIn $\Delta ABC$,$AD$ is the perpendicular bisector of $BC$ (see figure). Show that $\Delta ABC$ is an isosceles triangle in which $AB = AC$.
Medium
View SolutionIn the figure,$\angle B < \angle A$ and $\angle C < \angle D$. Show that $AD < BC$.
Difficult
View SolutionIn right triangle $ABC$,right-angled at $C$,$M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$ (see Fig). Show that:
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$
Difficult
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