*Q1) In a parallelogram ABCD, determine the sum of angles \(\angle C\;and\;\angle D\).*

**Solution:**

\(\angle C\;and\;\angle D\)

\(âˆ´ \angle C+\angle D=180^{0}\)

*Q2) In a parallelogram ABCD, if \(\angle B=135^{0}\), determine the measures of its other angles.*

**Solution:**

Given \(\angle B=135^{0}\)

ABCD is a parallelogram

\(âˆ´ \angle A=\angle C,\;\angle B=\angle D\;and\;\angle A+\angle B=180^{0}\)

\(\Rightarrow \angle A+135^{0}=180^{0}\)

\(\Rightarrow \angle A=45^{0}\)

\(\Rightarrow \angle A=\angle C=45^{0}\;and\;\angle B=\angle C=135^{0}\)

*Q3) ABCD is a square. AC and BD intersect at O. State the measure of \(\angle AOB\).*

**Solution:**

Since, diagonals of a square bisect each other at right angle.

\(âˆ´ \angle AOB=90^{0}\)

*Q4) ABCD is a rectangle with \(\angle ABD=40^{0}\). Determine \(\angle DBC\)*

**Solution:**

We have,

\(\angle ABC=90^{0}\)

\( \Rightarrow \angle ABD+\angle DBC=90^{0}\;\;\;\;\;\;\;\;[âˆµ \angle ABD=40^{0}]\)

\(\Rightarrow 40^{0}+\angle DBC=90^{0}\)

\(âˆ´ \angle DBC=50^{0}\)

*Q5) The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.*

**Solution:**

Since ABCD is a parallelogram

\(âˆ´ AB\parallel DC\;and\;AB=DC\)

\(\Rightarrow EB\parallel DF\;and\;\;\frac{1}{2}AB=\frac{1}{2}DC\)

\(\Rightarrow EB\parallel DF\;and\;EB=DF\)

EBFD is a parallelogram.

*Q6) P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.*

**Solution:**

We know that,

Diagonals of a parallelogram bisect each other.

Therefore, OA = OC and OB = OD

Since P and Q are point of intersection of BD.

Therefore, BP = PQ = QD

Now, OB = OD are BP = QD

=>OB â€“ BP = OD â€“ QD

=>OP = OQ

Thus in quadrilateral APCQ, we have

OA = OC and OP = OQ

Diagonals of Quadrilateral APCQ bisect each other.

Therefore APCQ is a parallelogram.

Hence \(AP\parallel CQ\)

*Q7) ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.*

**Solution:**

We have,

AE = BF = CG = DH = x (say)

BE = CF = DG = AH = y (say)

\(In\;\Delta AEH\;and\Delta BEF,\;we\;have\)

AE = BF

\(\angle A=\angle B\)

And AH = BE

So, by SAS congruency criterion, we have

\(\Delta AEH\simeq \Delta BFE\)

\(\Rightarrow \angle 1=\angle 2\;and\;\angle 3=\angle 4\)

But \(\angle 1+\angle 3=90^{0}\;and\;\angle 2+\angle A=90^{0}\)

\(\Rightarrow \angle 1+\angle 3+\angle 2+\angle A=90^{0}+90^{0}\)

\(\Rightarrow \angle 1+\angle 4+\angle 1+\angle 4=180^{0}\)

\(\Rightarrow 2(\angle 1+\angle 4)=180^{0}\)

\(\Rightarrow \angle 1+\angle 4=90^{0}\)

HEF=90^{0}

Similarly we have \(\angle F=\angle G=\angle H=90^{0}\)

Hence, EFGH is a Square.

*Â *

*Q8) ABCD is a rhombus, EAFB is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.*

**Solution:**

We know that the diagonals of a rhombus are perpendicular bisector of each other.

\(âˆ´ OA=OC,\;OB=OD,\;and\;\angle AOD=\angle COD=90^{0}\)

And \(\angle AOB=\angle COB=90^{0}\)

In \(\Delta BDE\)

\(OA\parallel DE\)

\(OC\parallel DG\)

In \(\Delta CFA\)

\(OB\parallel CF\)

\(OD\parallel GC\)

Thus, in quadrilateral DOGC, we have

\(OC\parallel DG\;and\;OD\parallel GC\)

=>DOCG is a parallelogram

\(\angle DGC=\angle DOC\)

\(\angle DGC=90^{0}\)

*Q9) 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.*

**Solution:**

Draw a parallelogram ABCD with AC and BD intersecting at O.

Produce AD to E such that DE = DC

Join EC and produce it to meet AB produced at F.

In \(\Delta DCE\)

\(\angle DCE=\angle DEC…..(i)\)

\(AB\parallel CD\)

\( âˆ´ AE\parallel CD\)

\(AF\parallel CD\)

\(\angle DCE=\angle BFC…..(ii)\)

From (i) and (ii) we get

\(\angle DEC=\angle BFC\)

In \(\Delta AFE\)

\(\angle AFE=\angle AEF\)

Therefore, AE = AF Â Â Â Â Â Â Â Â Â Â Â Â Â Â [In a triangle, equal angles have equal sides opposite to them]

=>AD + DE = AB + BF

=>BC + AB = AB + BF Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Since, AD = BC, DE = CD and CD = AB, AB = DE]

=> BC = BF

Hence proved.