**1** **In quadrilateral $ACBD, AC = AD$ and $AB$ bisects $\angle A$ (See the given figure). Show that $\Delta ABC \cong \Delta ABD$. What can you say about $BC$ and $BD$?**

In $\Delta ABC$ and $\Delta ABD,$$\\$ $AC = AD$ (Given)$\\$ $\angle CAB = \angle DAB (AB$ bisects $\angle A)$$\\$ $AB = AB $(Common)$\\$ $\therefore \Delta ABC \cong \Delta ABD$ (By $SAS$ congruence rule)$\\$ $\therefore BC = BD (By \ CPCT)$$\\$ Therefore, $BC$ and $BD$ are of equal lengths.

**2** **$ABCD $ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (See the given figure). Prove that$\\$ (i) $\Delta ABD \cong \Delta BAC$$\\$ (ii)$ BD = AC$$\\$ (iii) $\angle ABD = \angle BAC.$**

In $\Delta ABD$ and $\Delta BAC,$$\\$ $AD = BC$ (Given)$\\$ $\angle DAB = \angle CBA $(Given)$\\$ $AB = BA$ (Common)$\\$ $\therefore \Delta ABD \cong \Delta BAC$ (By $SAS$ congruence rule)$\\$ $\therefore BD=Ac(By \ CPCT)$$\\$ And, $\angle ABD=\angle BAC(By\ CPCT) $

**3** **$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (See the given figure). Show that $CD$ bisects $AB$.**

In $\Delta BOC $ and $\Delta AOD,$$\\$ $\angle BOC = \angle AOD $(Vertically opposite angles)$\\$ $\angle CBO = \angle DAO$ (Each $90^o$)$\\$ $BC = AD$ (Given)$\\$ $\therefore \Delta BOC \cong \Delta AOD (AAS$ congruence rule)$\\$ $\therefore BO = AO (By \ CPCT)$$\\$ $ CD$ bisects $AB.$

**4** **$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see the given figure). Show that $\Delta ABC \cong \Delta CDA.$$\\$**

In $\Delta ABC $ and $\Delta CDA,$$\\$ $\angle BAC = \angle DCA$ (Alternate interior angles, as $p \parallel q$)$\\$ $AC = CA$ (Common)$\\$ $\angle BCA = \angle DAC $(Alternate interior angles, as $l \parallel m$)$\\$ $\therefore \Delta ABC \cong \Delta CDA$ (By $ASA$ congruence rule)

**5** **Line $l$ is the bisector of an angle $\angle A$ and $\angle B$ is any point on $l$. $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle A$ (see the given figure). Show that:$\\$ (i) $\Delta APB \cong \Delta AQB$$\\$ (ii)$ BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.**

In $\Delta APB$ and $\Delta AQB,$$\\$ $\angle APB = \angle AQB$ (Each $90^o)$$\\$ $\angle PAB = \angle QAB$ ($l$ is the angle bisector of $\angle A$)$\\$ $AB = AB$ (Common)$\\$ $\therefore \Delta APB \cong \Delta AQB$ (By $AAS$ congruence rule)$\\$ $\therefore BP = BQ$ (By $ CPCT)$$\\$ Or, it can be said that B is equidistant from the arms of $\angle A$.