- 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.
Data: ABCD is a parallelogram E and F is the mid-points of AD and BC
To prove: Area of parallelogram ABFE = area of | | gm FEDC
|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 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.
Given: in Δ ABC, D is the mid-point of AB. DE | | BC
To prove: E is the midpoint of AC
|1. In Δ ABC, D is the mid-point of AB. DE | | BC|
|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.
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.
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)
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.
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.
|In PQR, x, y, and z are the|
Mid-point of PQ QR and RP
|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|
- 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.
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
|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|
6. Prove that if the mid-points of the opposite sides of a quadrilateral are joined, they bisect each other.
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.
|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|