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 \displaystyle \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



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



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


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


Consider \triangle EDC and \triangle BFC

\angle EDC = \angle CBF

In \triangle AEF

Since DC \parallel AB

\displaystyle \frac{ED}{AD} + \frac{EC}{CF} … … … … … i)

Also since AD \parallel BC

\displaystyle \frac{CF}{EC} + \frac{BF}{AB} … … … … … i)

From i) and ii)

\displaystyle \frac{BF}{AB} + \frac{AD}{ED}

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