$ABC$ 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
$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
Difficult
View SolutionIn a triangle,locate a point in its interior which is equidistant from all the sides of the triangle.
Medium
View Solution$ABC$ is a right-angled triangle in which $\angle A = 90^o$ and $AB = AC$. Find $\angle B$ and $\angle C$.
Medium
View Solution$AB$ is a line segment and line $l$ is its perpendicular bisector. If a point $P$ lies on $l$,show that $P$ is equidistant from $A$ and $B$.
Medium
View SolutionLine-segment $AB$ is parallel to another line-segment $CD$. $O$ is the mid-point of $AD$ (see Fig). Show that $(i)$ $\Delta AOB \cong \Delta DOC$ $(ii)$ $O$ is also the mid-point of $BC$.
Medium
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