Class 9 Maths

## 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.

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

∠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

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