In an isosceles triangle ABC, with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that : (i) $OB = OC$

ABCD is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see Fig. 7.17). Prove that (i) $\triangle ABD \cong \triangle BAC$

Show that the angles of an equilateral triangle are $60^{\circ}$ each.

In an isosceles triangle ABC, with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that : (ii) $AO$ bisects $\angle A$

In 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. 7.23). Show that: (iv) $CM = \frac{1}{2} AB$

In $\triangle ABC$, $AD$ is the perpendicular bisector of $BC$ (see Fig. 7.30). Show that $\triangle ABC$ is an isosceles triangle in which $AB = AC$.

ABC is a right angled triangle in which $\angle A = 90^{\circ}$ and $AB = AC$. Find $\angle B$ and $\angle C$.

$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig. 7.19). Show that $\triangle ABC \cong \triangle CDA$.

$AB$ 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 $\angle EPA = \angle DPB$ (see Fig. 7.22). Show that (ii) $AD = BE$

$\triangle ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$ (see Fig. 7.34). Show that $\angle BCD$ is a right angle.

In quadrilateral ACBD, $AC = AD$ and $AB$ bisects $\angle A$ (see Fig. 7.16). Show that $\triangle ABC \cong \triangle ABD$. What can you say about $BC$ and $BD$?

$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see Fig. 7.18). Show that $CD$ bisects $AB$.

In 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. 7.23). Show that: (i) $\triangle AMC \cong \triangle BMD$

$\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig. 7.39). If $AD$ is extended to intersect $BC$ at $P$, show that (iii) $AP$ bisects $\angle A$ as well as $\angle D$.

ABCD is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see Fig. 7.17). Prove that (iii) $\angle ABD = \angle BAC$.

ABC is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see Fig. 7.32). Show that (i) $\triangle ABE \cong \triangle ACF$

In 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. 7.23). Show that: (ii) $\angle DBC$ is a right angle.

ABCD is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see Fig. 7.17). Prove that (ii) $BD = AC$

$\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig. 7.39). If $AD$ is extended to intersect $BC$ at $P$, show that (iv) $AP$ is the perpendicular bisector of $BC$.

ABC is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively (see Fig. 7.31). Show that these altitudes are equal.