Jitender Gupta and Divya Malik Solutions for Exercise 4: CHALLENGERS

$ABCD$ is a quadrilateral in which $AB=BC$. $E$ is the point of intersection of the perpendicular bisectors of $AD$ and $CD$. Then $\angle ABC$ is

If a point moves so that the sum of its square of the distances from two intersecting straight lines is always constant, then the locus of the moving point is

The locus of the vertex of a triangle of given base $AB$ and height $h$ is a pair of lines

In the given figure, $AX$ bisects $\angle BAC$ and $PQ$ is perpendicular bisector of $AC$, which meets $AX$ at point $Y$.

The following statements are:

$\mathrm{I}$. $X$ is equidistant from $AB$ and $AC$.

$\mathrm{II}$. $Y$ is equidistant from $A$ and $C$.

We construct a $△ABC$, in which $AB=4.2 \mathrm{cm}, BC=6.3 \mathrm{cm}$ and $AC=5 \mathrm{cm}$, and draw perpendicular bisector of $BC$, which meets $AC$ at point $D$. The point $D$ is equidistant from

We construct a right angled $△PQR$, in which $\angle Q=90°$, hypotenuse $PR=8 \mathrm{cm}$ and $QR=4.5 \mathrm{cm}$, and draw bisector of $\angle PQR$ which meets $PR$ at point $T$. Then,

We construct a $△ABC$, in which $\angle ABC=75°, AB=5 \mathrm{cm}$ and $BC=6.4 \mathrm{cm}$. Draw perpendicular bisector of side $BC$ and also bisector of $\angle ACB$. If these bisectors intersect each other at point $P$. Which of the following statements is/are true?

$\mathrm{I}$. $P$ is equidistant from $B$ and $C$.

$\mathrm{II}$. $P$ is equidistant from $AC$ and $BC$.