 Prove the statement of example 4 using the following ideas: Join DF and product it to meet AB produced in G. Show the two parallelograms ABFE and EFCD have equal area using the converse of midpoint theorem.
Solution:
Data: ABCD is a parallelogram E and F is the midpoints of AD and BC
Respectively.
To prove: Area of parallelogram ABFE = area of   gm FEDC
Proof:  
In Δ DCF and Δ BFG,  
∠DCF = ∠FBG  (alternate angles, DC   AB)

∠DFC = ∠BFG  (V.O.A)

FC = FB  (F is midpoint)

⸫ Δ DCF ≈ Δ BFC  (ASA postulate) 
⸫ DC = BG
 
But DC = AB
 
⸫ AB = BG
 
Now area of parallelogram ABEF = 2 area of BGF (1)

(∵Parallelogram and Δ standing on equal base and between same parallel)
Now Area of parallelogram EDCF = 2 area of Δ DCF
But Δ DCF ≈ Δ BGFS
⸫ Area of parallelogram EDCF = 2 area of Δ BGF (2)
From (1) and (2)
Area of parallelogram ABEF = Area of parallelogram EDCF
 Prove the converse of the midpoint theorem following the guidelines given below.
Consider a triangle ABC with D as the minpoint of AB. Draw DE   BC to intersect AC in E. let E1 be the midpoint of ac. Use midpoint theorem to get DE1   BC and DE1 = 𝐁𝐂/2 . Conclude E = E1 and hence E is the midpoint of AC.
Solution:
Given: in Δ ABC, D is the midpoint of AB. DE   BC
To prove: E is the midpoint of AC
Proof:  
Statement  Reason

1. In Δ ABC, D is the midpoint of AB. DE   BC
 Given

Let E, be the midpoint of AC. Join DE, D is the midpoint of AB and E1 is the midpoint of AC ∴ DE1   BC  Midpoint theorem parallels.

But DE   BC  Given

This is possible only if E and E1 consider (Through a given point, only one line can be drawn   to be a given line) ∴ E and E1 coincide i.e. DE1 is the same as DE Thus a line drawn through the midpoint of a side of a triangle and parallel to another bisects the third side. 
3. In a rectangle ABCD, P, Q, R and S are the midpoint of the sides AB, BC, CD and DA respectively. Find the area of PQRS in terms of area of ABCD.
Solution:
Given: ABCD is a rectangle, P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.
To find: the areas (PQRS) in terms of area (ABCD)
Proof: S and Q are the midpoint of AD and BC
∴ SQ divides rectangles ABCD into two rectangles equal in area.
Area (ABQS) = area (SQCD) = ^{1}/_{2} area (ABCD)
Δ SRQ and parallelogram (rectangles) SQCD stand on the same base SQ and are between the same parallel.
∴ Area Δ SRQ = ^{1}/_{2} area (SQCD)
Similarly area Δ SPQ = ^{1}/_{2} area (SABQ)
Adding, area Δ SRQ + area Δ SPQ = ^{1}/_{2} area (SQCD) + ^{1}/_{2} area (SABQ)
i.e., area (PQRS) = ^{1}/_{2} area (ABCD)
4. Suppose x, y and z are the midpoint of the sides PQ, QR and RP. Respectively of a triangle PQR. Prove that XYPZ is a parallelogram.
Solution:
Given: x, y and z are the midpoint of the sides PQ, QR and RP of Δ PQR, xy and xz are joined.
Prove: XYRZ is a parallelogram.
Proof:  
Statement  Reason

In PQR, x, y, and z are the Midpoint of PQ QR and RP Respectively.
 
XY   PR  midpoint theorem

XY = ^{1}/_{2} PR  midpoint theorem

ZR = ^{1}/_{2} PR  Given

Therefore, XY RZ is a Parallelogram.
 one pair of opposite sides parallel and equal 
 Suppose EFGH is a trapezium in which EF is parallel to HG. Through X, the midpoint EH, XY is drawn parallel to EF meeting FG at Y. prove that XY bisects FG.
Solution:
Given: EFGH is a trapezium with EF   HG. Through X the midpoint of EH,
Is drawn parallel to EF
To prove: XY bisects FG
Construction: Join HF to intersect XY at P
Proof  
Statement  Reason

In Δ HEF XP   EF  converse of midpoint theorem

∴ XP bisects HF PY   HG
 HY   EF and EF   HG

Δ HGF, PY passes through midpoint of HF and is parallel to HG ∴ PY bisects FG  converse of midpoint theorem

6. Prove that if the midpoints of the opposite sides of a quadrilateral are joined, they bisect each other.
Solution:
Data: ABCD is a quadrilateral P, Q, R and S are midpoints of AB, BC, CD and DA respectively.
To prove: PR and SQ bisect each other.
Construction: join AC.
Proof:  
Statement  Reason

In Δ ABC, PQ   AC, and PQ = 12 AC  midpoint theorem ……. (1)

In Δ ADC, SR   AC, and SR = 12 AC  midpoint theorem ……. (2)

∴ PQ   SR and PQ = SR  from (1) and (2) opposite sides equal and parallel

∴PQRS is a parallelogram PR and SQ are diagonals PR bisect SQ
 
One thought on “THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.4 – Class IX”
Comments are closed.