**1** **The angles of quadrilateral are in the ratio $3:5:9:13$. Find all the angles of the quadrilateral.**

Let the common ratio between the angles be $x.$$\\$ Therefore, the angles will be $3x, 5x, 9x,$ and $13x$ respectively.$\\$ As the sum of all interior angles of a quadrilateral is $360^o,$$\\$ $\therefore 3x+5x+9x+13x=360^o\\ 30x =360^o\\ x=12^o$$\\$ Hence, the angles are$\\$ $3x=3*12=36^o\\ 5x=5*12=60^o\\ 9x=9*12=108^o\\ 13x=13*12=156^o$

**2** **If the diagonals of a parallelogram are equal, then show that it is a rectangle.**

Let $ABCD$ be a parallelogram. To show that $ABCD$ is a rectangle, we have to prove that one of its interior angles is $90^o$.$\\$ In $\Delta ABC$ and $\Delta DCB,$$\\$ $AB = DC$ (Opposite sides of a parallelogram are equal)$\\$ $BC = BC$ (Common)$\\$ $AC = DB$ (Given)$\\$ $\therefore \Delta ABC \cong \Delta DCB$ (By $SSS$ Congruence rule)$\\$ $\Rightarrow \angle ABC = \angle DCB$$\\$ It is known that the sum of the measures of angles on the same side of transversal is $180^o$.$\\$ $\angle ABC + \angle DCB = 180^o (AB \parallel CD)$$\\$ $\Rightarrow ABC + \angle ABC = 180^o$$\\$ $\Rightarrow 2\angle ABC = 180^o$$\\$ $\Rightarrow ABC = 90^o$$\\$ Since $ABCD$ is a parallelogram and one of its interior angles is $90^o$, $ABCD$ is a rectangle.

**3** **Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.**

Let $ABCD$ be a quadrilateral, whose diagonals $AC$ and $BD$ bisect each other at right angle i.e. $OA = OC, OB = OD,$ and $\angle AOB = \angle BOC = \angle COD = \angle AOD = 90^o.$$\\$ To prove $ABCD$ a rhombus,$\\$ We have to prove $ABCD$ is a parallelogram and all the sides of $ABCD$ are equal.$\\$ In $\Delta AOD$ and $\Delta COD,$$\\$ $OA = OC$ (Diagonals bisect each other)$\\$ $\angle AOD = \angle COD$ (Given)$\\$ $OD = OD$ (Common)$\\$ $\therefore \Delta AOD \cong \Delta COD$ (By $SAS$ congruence rule)$\\$ $\therefore AD = CD$ ... (1)$\\$ Similarly, it can be proved that$\\$ $AD = AB$ and $CD = BC$ ... (2)$\\$ From Equations (1) and (2),$\\$ $AB = BC = CD = AD$$\\$ Since opposite sides of quadrilateral $ABCD$ are equal, it can be said that $ABCD$ is a parallelogram. Since all sides of a parallelogram $ABCD$ are equal, it can be said that $ABCD$ is a rhombus.

**4** **Show that the diagonals of a square are equal and bisect each other at right angles.**

Let $ABCD$ be a square.$\\$ Let the diagonals $AC$ and $BD$ intersect each other at a point $O.$$\\$ To prove that the diagonals of a square are equal and bisect each other at right angles,$\\$ we have to prove,$\\$ $AC = BD, OA = OC, OB = OD,$ and $\angle AOB = 90^o.$$\\$ In $\Delta ABC$ and $\Delta DCB$,$\\$ $AB = DC $(Sides of a square are equal to each other)$\\$ $\angle ABC = \angle DCB$ (All interior angles are of $90^o$ )$\\$ $BC = CB$ (Common side)$\\$ $\therefore \Delta ABC \cong \Delta DCB $(By $SAS$ congruency)$\\$ $\therefore AC = DB$ (By $ CPCT$)$\\$ Hence, the diagonals of a square are equal in length.$\\$ In $\Delta AOB$ and $\Delta COD,$$\\$ $\angle AOB = \angle COD $(Vertically opposite angles)$\\$ $\angle ABO = \angle CDO$ (Alternate interior angles)$\\$ $AB = CD$ (Sides of a square are always equal)$\\$ $\therefore \Delta AOB \cong \Delta COD$ (By $AAS$ congruence rule)$\\$ $\therefore AO = CO$ and $OB = OD $(By $CPCT$)$\\$ Hence, the diagonals of a square bisect each other.$\\$ In $\Delta AOB$ and $\Delta COB$,$\\$ As we had proved that diagonals bisect each other, therefore,$\\$ $AO = CO$$\\$ $AB = CB$ (Sides of a square are equal)$\\$ $BO = BO$ (Common)$\\$ $\therefore \Delta AOB \cong \Delta COB $ (By $SSS$ congruency)$\\$ $\therefore \angle AOB = \angle COB$ (By $CPCT$)$\\$ However, $\angle AOB + \angle COB = 180^o$ (Linear pair)$\\$ $2\angle AOB = 180^o$$\\$ $\angle AOB = 90^o$$\\$ Hence, the diagonals of a square bisect each other at right angles.

**5** **Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.**

Let us consider a quadrilateral $ABCD$ in which the diagonals $AC$ and $BD$ intersect each other at $O$.$\\$ It is given that the diagonals of $ABCD$ are equal and bisect each other at right angles.$\\$ Therefore, $AC = BD, OA = OC, OB = OD, $ and $\angle AOB = \angle BOC = \angle COD = \angle AOD = 90^o$.$\\$ To prove $ABCD$ is a square$\\$ We have to prove that $ABCD$ is a parallelogram, $AB = BC = CD = AD,$ and one of its interior angles is$90^o$.$\\$ In $\Delta AOB $ and $\Delta COD,$$\\$ $AO = CO$ (Diagonals bisect each other)$\\$ $OB = OD$ (Diagonals bisect each other)$\\$ $\angle AOB = \angle COD $(Vertically opposite angles)$\\$ $\therefore \Delta AOB \cong \Delta COD$ (SAS congruence rule)$\\$ $\therefore AB = CD $(By $CPCT)$$\\$ ... (1) And, $\angle OAB = \angle OCD $(By $CPCT)$$\\$ However, these are alternate interior angles for line $AB$ and $CD$ and alternate interior angles are equal to each other only when the two lines are parallel.$\\$ $AB \parallel CD ... (2)$$\\$

From Equations (1) and (2), we obtain$\\$ $ABCD $ is a parallelogram.$\\$ In $\Delta AOD $ and $\Delta COD,$$\\$ $AO = CO$ (Diagonals bisect each other)$\\$ $\angle AOD = \angle COD$ (Given that each is $90^o$)$\\$ $OD = OD$ (Common)$\\$ $\therefore \Delta AOD \cong \Delta COD$ ($SAS$ congruence rule)$\\$ $\therefore AD = DC ... (3)$$\\$ However,$ AD = BC$ and $AB = CD $ (Opposite sides of parallelogram $ABCD$)$\\$ $AB = BC = CD = DA$$\\$ Therefore, all the sides of quadrilateral $ABCD$ are equal to each other.$\\$ In $\Delta ADC$ and $\Delta BCD$,$\\$ $AD = BC$ (Already proved)$\\$ $AC = BD $(Given)$\\$ $DC = CD $(Common)$\\$ $\therefore \Delta ADC \cong \Delta BCD$ ($SSS$ Congruence rule)$\\$ $\therefore \angle ADC = \angle BCD$ (By $CPCT)$$\\$ However, $\angle ADC + \angle BCD = 180^o$ (Co-interior angles)$\\$ $\angle ADC + \angle ADC = 180^o$$\\$ $2\angle ADC = 180^o$$\\$ $\therefore \angle ADC = 90^o$$\\$ One of the interior angles of quadrilateral $ABCD $ is a right angle.$\\$ Thus, we have obtained that $ABCD$ is a parallelogram, $AB = BC = CD = AD$and one of its interior angles is $90^o.$$\\$ Therefore, $ABCD$ is a square.