(a) In the quadrilateral (1) given below, AD = BC, P, Q, R, and S are mid-points of AB, BD, CD, and AC
respectively. Prove that PQRS is a rhombus.
(b) In the figure (2) given below, ABCD is a kite in which BC = CD, AB = AD, E, F, G are mid-points of
CD, BC, and AB respectively. Prove that:
(i) ∠EFG = 90^{\circ}
(ii) The line drawn through G and parallel to FE bisects DA.
(a) It is given that
A quadrilateral ABCD in which AD = C
P, Q, R, and S are mid-points of AB, BD, CD, and AC
To prove:
PQRS is a rhombus
Proof:
In Δ ABD
P and Q are mid points of AB and BD
PQ || AD and PQ = ½ AB …. (1)
In Δ BCD,
R and Q are mid points of DC and BD
RQ || BC and RQ = ½ BC …… (2)
P and S are mid-points of AB and AC
PS || BC and PS = ½ BC ……. (3)
AD = BC
Using all the equations
PS || RQ and PQ = PS = RQ
Here PS || RQ and PS = RQ
PQRS is a parallelogram
PQ = RS = PS = RQ
PQRS is a rhombus
Therefore, it is proved
(ii) It is given that
ABCD is a kite in which BC = CD, AB = AD, E, F, G are mid-points of CD, BC, and AB
To prove:
(i) ∠EFG = 90^{\circ}
(ii) The line drawn through G and parallel to FE bisects DA
Construction:
Join AC and BD
Construct GH through G parallel to FE
Proof:
(i) We know that
Diagonals of a kite interest at right angles
∠MON = 90^{\circ} .......(1)
In Δ BCD
E and F are mid-points of CD and BC
EF || DB and EF = ½ DB …… (2)
EF || DB
MF || ON
Here
∠MON + ∠MFN = 180^{\circ}
90^{\circ} + ∠MFN = 180^{\circ}
By further calculation
∠MFN = 180 – 90 = 90^{\circ}
Hence, it is proved.
(ii) In Δ ABD
G is the mid-point of AB and HG || DB
Using equation (2)
EF || DB and EF || HG
HG || DB
Here H is the mid-point of DA
Therefore, the line drawn through G and parallel to FE bisects DA.
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies