# Quadrilaterals

## Exercise 8.1 Part 2

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

Answer: Using the same figure,

If DO=AO

Then ∠DAO=∠BAO=45°

(Angles opposite to equal sides are equal)
So, all angles of the quadrilateral are right angles making it a square.

Question 6: Diagonal AC of a parallelogram ABCD bisects angle A . Show that

(i) it bisects angle C also,
(ii) ABCD is a rhombus.

Answer: ABCD is a parallelogram where diagonal AC bisects angle DAB

In ΔADC and ΔABC
∠DAC=∠BAC (diagonal is bisecting the angle)
AC=AC (common side)
AD=BC (parallel sides are equal in a parallelogram)
Hence, ΔADC≅ΔABC
So, ∠DCA=∠BCA
This proves that AC bisects ∠DCB as well.

Now let us assume another diagonal DB intersecting AC on O.
As it is a parallelogram so DB will bisect AC and vice-versa.
In ΔAOD and ΔBOD
∠DAO=∠DCO
(opposite angles are equal in parallelogram so their halves will be equal)
AO=CO
DO=DO
Hence ΔAOD ≅ ΔBOD
So, ∠DOA=∠DOB=90°

As diagonals are intersecting at right angles so it is a rhombus

Question 7: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that:

1. ΔAPD≅ ΔCQB
2. AP=CQ
3. ΔAQB≅ΔCPD
4. AQ=CP
5. APCQ is a parallelogram

Answer: In ΔAPD and ΔCQB
DP=BQ (given)
AD=BC (opposite sides are equal)
∠DAP=∠BCQ (opposite angles’ halves are equal)
Hence, ΔAPD≅ΔCQB
So, AP=CQ proved

In ΔAQB and ΔCPD
AB=CD (opposite sides are equal)
DP=BQ (given)
∠BAQ=∠DCP (opposite angles’ halves are equal)
Hence, ΔAQB≅ΔCPD
So, AQ=CP proved
∠DPA=∠BQP
(corresponding angles of congruent triangles APD and CQB)

In ΔDQP and ΔBQP
∠DPQ=∠BQP
(from previous proof)
DP=BQ (given)
PQ=PQ (common side)
So, ΔDQP≅ΔBQP
So, ∠QDP=∠QBP

With equal opposite angles and equal opposite sides it is proved that APCQ is a parallelogram

Question 8: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that

1. ΔAPB ≅ ΔCQD
2. AP = CQ

Answer: In ΔAPB and ΔCQD
∠ABP=∠CDQ (alternate angles of transversal DB)
AB=CD
∠APB=∠CQD (right angles)
Hence, ΔAPB≅ΔCQD
So, AP=CQ

Question 9: In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively. Show that

1. quadrilateral ABED is a parallelogram
2. quadrilateral BEFC is a parallelogram
3. AD || CF and AD = CF
4. quadrilateral ACFD is a parallelogram
5. AC = DF
6. ΔABC ≅ ΔDEF

Answer: In ΔABC and ΔDEF
AB=DE (given)
BC=EF (given)
∠ABC=∠DEF (AB||DE and BC||EF)
Hence, ΔABC≅ΔDEF

In quadrilateral ABED
AB= ED
AB||ED
So, ABED is a parallelogram (opposite sides are equal and parallel)
So, BE||AD ------------ (1)
Similarly quadrilateral ACFD can be proven to be a parallelogram
So, BE||CF ------------ (2)
From equations (1) & (2)
It is proved that
AD||CF
So, AD=CF
Similarly AC=DF and AC||DF can be proved

Question 10: ABCD is a trapezium in which AB || CD and AD = BC. Show that

1. ∠A=∠B
2. ∠C=∠D
3. ΔABC ≅ ΔBAD
4. Diagonal AC = Diagonal BD

Answer: In ΔBCE EC=AD (opposite sides are equal in parallelogram)
AD=BC (given)
So, BC=EC
So, ∠CBE=∠CEB
∠CBE+∠CBA=180° (linear pair of angles)
∠CEB+∠DAB=180° (adjacent angles of parallelogram are complementary)
substituting ∠CBE=∠CEB it is clear that ∠DBA=∠CBA

Now, ∠DAB+∠CDA=180° (adjacent angles of parallelogram)
And, ∠CBA+∠DCB=180° (adjacent angles of parallelogram)
As ∠DBA=∠CBA so it is clear that ∠CDA=∠DCB

In ΔABC and ΔBAD
AB=AB (common side)
AD=BC (given)
∠DBA=∠CBA
hence, ΔABC≅ΔBAD

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