1. 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 mid-point theorem.

Solution:

Data: ABCD is a parallelogram E and F is the mid-points 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

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2. Prove the converse of the mid-point theorem following the guidelines given below.

Consider a triangle ABC with D as the min-point of AB. Draw DE | | BC to intersect AC in E. let E1 be the mid-point of ac. Use mid-point theorem to get DE1 | | BC and DE1 = 𝐁𝐂/2 . Conclude E = E1 and hence E is the mid-point of AC.

Solution:

Given: in Δ ABC, D is the mid-point of AB. DE | | BC

To prove: E is the midpoint of AC

Proof:
Statement	Reason
1. In Δ ABC, D is the mid-point of AB. DE | | BC	Given
Let E, be the mid-point of AC. Join DE, D is the mid-point of AB and E1 is the mid-point of AC ∴ DE1 | | BC	Mid-point 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 mid-point of a side of a triangle and parallel to another bisects the third side.

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3. In a rectangle ABCD, P, Q, R and S are the mid-point 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 mid-point 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)

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4. Suppose x, y and z are the mid-point 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 mid-point 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 Mid-point of PQ QR and RP Respectively.
XY | | PR	mid-point theorem
XY = 1/2 PR	mid-point theorem
ZR = 1/2 PR	Given
Therefore, XY RZ is a Parallelogram.	one pair of opposite sides parallel and equal

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5. Suppose EFGH is a trapezium in which EF is parallel to HG. Through X, the mid-point 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 mid-point 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 mid-point theorem
∴ XP bisects HF PY | | HG	HY | | EF and EF | | HG
Δ HGF, PY passes through mid-point of HF and is parallel to HG ∴ PY bisects FG	converse of mid-point theorem

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6. Prove that if the mid-points 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 mid-points 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	mid-point theorem ……. (1)
In Δ ADC, SR | | AC, and SR = 12 AC	mid-point 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

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