Prove that bisectors of any two adjacent angles of a parallelogram are at right angles

Answer

Note: A property of that parallelogram says that if a parallelogram has all sides equal, then their diagonal bisector intersects perpendicularly. An angle bisector is a ray that splits an angle into two consecutive congruent, smaller angles.

In ∆ABC and ∆DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, Band C are joined to vertices D, E and F respectively (see figure). Show that:(i)     quadrilateral ABED is a parallelogram(ii)    quadrilateral BEFC is a parallelogram(iii)   AD || CF and AD = CF(iv)   quadrilateral ACFD is a parallelogram

(v)     AC = DF
(vi)    ∆ABC ≅ ∆DEF. [CBSE 2012

(i) Given AM bisect angle A and BM bisects angle of || gm ABCD.

Hence, bisectors of any two adjacent angles of a parallelogram are at right angles.

(ii) Given: A || gm ABCD in which bisector AR of ∠A meets DC in R and bisector CQ of ∠C meets AB

in Q

To prove: AR || CQ

Proof:

In || gm ABCD, we have

∠A = ∠C [Opposite angles of || gm are equal]

½ ∠A = ½ ∠C

∠DAR = ∠BCQ [Since, AR is bisector of ½ ∠A and CQ

is the bisector of ½ ∠C]

Now, in ∆ADR and ∆CBQ

∠DAR = ∠BCQ [Proved above]

AD = BC [Opposite sides of || gm ABCD are

equal]

So, ∆ADR ≅ ∆CBQ, by A.S.A axiom of congruency

Then by C.P.C.T, we have

∠DRA = ∠BCQ

And,

∠DRA = ∠RAQ [Alternate angles since, DC || AB]

Thus, ∠RAQ = ∠BCQ

But these are corresponding angles,

Hence, AR || CQ.

(iii) Given: In quadrilateral ABCD, diagonals AC and BD are equal and bisect each other at right angles

To prove: ABCD is a square

Proof:

In ∆AOB and ∆COD, we have

AO = OC [Given]

BO = OD [Given]

∠AOB = ∠COD [Vertically opposite angles]

So, ∆AOB ≅ ∆COD, by S.A.S axiom of congruency

By C.P.C.T, we have

AB = CD

and ∠OAB = ∠OCD

But these are alternate angles

AB || CD

Thus, ABCD is a parallelogram

In a parallelogram, the diagonal bisect each other and are equal

Hence, ABCD is a square.