Question 1: In a parallelogram $ABCD$, determine the sum of angles $\angle A$ and $\angle B$.

Since $ABCD$ is a parallelogram, therefore $AD \parallel BC$

Since, $AD \parallel BC$ and transversal $AB$ intersects them at $A$ and $B$ respectively.

Since the sum of the interior angles on the same side of the transversal is $180^o$

Therefore $\angle A + \angle B = 180^o$

$\\$

Question 2: In a parallelogram $ABCD$, if $\angle B = 135^o$, determine the measures of its other angles.

Given $\angle B = 135^o$

Also $ABCD$ is a parallelogram.

Therefore $\angle A = \angle C$ and $\angle B = \angle D$ and $\angle A + \angle B = 180^o$

$\Rightarrow \angle A = 180^o - 135^o =45^o$

$\Rightarrow \angle C = 45^o$

$\Rightarrow \angle D = 180^o - 45^o =135^o$

$\\$

Question 3: $ABCD$ is a square. $AC$ and $BD$ intersect at $O$. State the measure of $\angle AOB$.

Consider $\triangle AOB$ and $\triangle AOD$

$AO$ is common

$AD = AB$ (side of a square)

$OB = OD$ (diagonals of a square bisect each other)

Therefore $\triangle AOB \cong \triangle AOD$

Hence $\angle AOD = \angle AOB$

Since $\angle AOD + \angle AOB = 180$

$\Rightarrow \angle AOD = 90^o$

$\\$

Question 4: $ABCD$ is a rectangle with $\angle ABD =40^o$. Determine $\angle DBC$.

Given $ABCD$ is a rectangle with $\angle ABD =40^o$

$\angle ABD + \angle DBC = 90^o$

$\Rightarrow \angle DBC = 90^o-40^o = 50^o$

$\\$

Question 5: The sides $AB$ and, $CD$ of a parallelogram $ABCD$ are bisected at $E$ and $F$. Prove that $EBFD$ is a parallelogram.

Given: $ABCD$ is a parallelogram, $E$ bisects $AB$ and $F$ bisects $DC$

Therefore $AB \parallel DC$ and $AB = DC$

$\Rightarrow EB \parallel DF$ also $\frac{1}{2} AB = \frac{1}{2} DC$

$\Rightarrow EB \parallel DF$ and $EB = DF$

Therefore $EBFD$ is a parallelogram.

$\\$

Question 6: $P$ and $Q$ are the points of trisection of the diagonal $BD$ of parallelogram $ABCD$. Prove $CQ$ is parallel to $AP$. Prove also that $AC$ bisects $PQ$.

Given: $DP = PQ = QB$

To prove: i) $CQ \parallel AP$ ii) $AC$ bisects $PQ$

Since diagonals of a parallelogram bisects each other

$OA = OC$ and $OB = OD$ … … … … … i)

Given $DP = PQ = QB$  … … … … … ii)

From i) and ii) we get

$OB - QB = OD - DP$

$\Rightarrow OQ = OP$

Now consider parallelogram $AQCP$

$OA = OC$

$OP = OQ$

We know that diagonals of a parallelogram bisects each other

Hence $AQCP$ is a parallelogram

Therefore $AP \parallel QC$

$\\$

Question 7: $ABCD$ is a square $E, F, G$ and $H$ are points on $BC, CD$ and $DA$ respectively, such that $AE = BF = CG = DH$. Prove that $EFGH$ is a square.

Given: $AE = BF = GC = DH$

Since $ABCD$ is a square  $\Rightarrow EB = CF = DG = AH$

Now consider $\triangle AEF$ and $\triangle BEF$

$\angle A = \angle B = 90^o$

$AE = BF$

$AH = EB$

Therefore $\triangle AEF \cong \triangle BEF$

$\Rightarrow EF = EH$

Therefore $\angle 1 = \angle 2$ and $\angle 3 = \angle 4$

$\Rightarrow \angle 2 + \angle 4 = 90^o$

$\Rightarrow \angle 1 + \angle 4 = 90^o$

$\Rightarrow \angle HEF = 90^o$

Similarly, we can prove that

$\angle EFG = \angle GHE = \angle EGH = 90^o$

Therefore $EFGH$ is a square

$\Rightarrow EFGH$ is a parallelogram

$\\$

Question 8: $ABCD$ is a rhombus, $EABF$ is a straight line such that $EA = AB = BF$. Prove that $ED$ and $FC$ when produced meet at right angles.

To prove: $\angle EGH = 90^o$

We know that the diagonals of a rhombus bisect each other and are perpendicular to each other.

Given: $OA = OC$ and $OB = OD$

$\angle AOD = \angle DOC = \angle COB = \angle BOA = 90^o$

$EA = AB =BF$

$AB = BC = CD = DA$ (sides of a rhombus)

In $\triangle BDE$, since $A$ is midpoint of $EB$ and $O$ is midpoint of $BD$

The line joining the midpoints i.e $AO \parallel ED \Rightarrow EDO = 90^o$

Similarly $BO \parallel FC \Rightarrow \angle FCO = 90^o$

Therefore in parallelogram $DOCG, OD \parallel CG$ and $DG \parallel OC$

Therefore $\angle DOC + \angle DGC = 180^o$

$\Rightarrow \angle DGC = 180^o - 90^o = 90^o$

$\Rightarrow \angle EGF = 90^o$

Hence proved.

$\\$

Question 9: $ABCD$ is a parallelogram, $AD$ is produced to $E$ so that $DE = DC$ and $EC$ produced meets $AB$ produced in $F$. Prove that $BF = BC$.

To prove: $BF = BC$

Given: $ED = DC = AB$

$AD = BC$

Consider $\triangle EDC$ and $\triangle BFC$

$\angle EDC = \angle CBF$

In $\triangle AEF$

Since $DC \parallel AB$

$\frac{ED}{AD}$ $=$ $\frac{EC}{CF}$ … … … … … i)

Also since $AD \parallel BC$

$\frac{CF}{EC}$ $=$ $\frac{BF}{AB}$ … … … … … i)

From i) and ii)

$\frac{BF}{AB}$ $=$ $\frac{AD}{ED}$

Therefore $BF = AD$ or $BF = BC$. Hence proved.

$\\$