In a huge park,people are concentrated at three points (see Fig) :
$A$ : where there are different slides and swings for children,
$B$ : near which a man-made lake is situated,
$C$ : which is near to a large parking and exit. Where should an ice cream parlour be set up so that the maximum number of persons can approach it?
(Hint: The parlour should be equidistant from $A$,$B$,and $C$.)

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(N/A) To ensure the ice cream parlour is equidistant from points $A$,$B$,and $C$,we need to find the circumcenter of the triangle formed by these points.
$1$. Join points $A$ and $B$,and draw line $l$,which is the perpendicular bisector of segment $AB$.
$2$. Join points $B$ and $C$,and draw line $m$,which is the perpendicular bisector of segment $BC$.
$3$. Let the perpendicular bisectors $l$ and $m$ intersect at point $O$.
Point $O$ is the circumcenter of $\triangle ABC$,and it is equidistant from vertices $A$,$B$,and $C$. Therefore,the ice cream parlour should be set up at point $O$.

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$ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see Fig). Show that
$(i)$ $\Delta ABE \cong \Delta ACF$
$(ii)$ $AB = AC$,i.e.,$ABC$ is an isosceles triangle.

$BE$ and $CF$ are two equal altitudes of a triangle $ABC$. Using $RHS$ congruence rule,prove that the triangle $ABC$ is isosceles.

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$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ bisects $\angle A$.

$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$

In the figure,$OA = OB$ and $OD = OC$. Show that
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.

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