Question

prove that the segment joining midpoints of non parallel sides of a trapezium is paralle to the parallel side and is half the sum of parralel sides

Answer 1

Answer:

Data: In the trapezium ABCD,

AD || BC, AX = XB and DY = YC

To Prove: (i) XY || AD     or    XY || BC

(ii) XY = 21(AD + BC)

Construction: Extend BA and CD to meet at Z.

Join A and C. Let it cut XY at P

Proof: (i)     In △ZBC,AD∣∣BC      [∵ Data]

∴ABZA=DCZD      [∵ BPT]

∴2AXZA=2DYZD         [∵ X & Y are mid points of AB and DC]

∴AXZA=DYZD

⇒XY∣∣AD           [∵ Converse of B.P.T.]

(ii) In △ABC,AX=XB       [∵ Data]

XP∣∣BC          [∵ Proved]

∴AP=PC        [∵ Converse of mid point theorem]

∴XP=21BC       [∵ Midpoint Theorem]

In △ADC,PY=21AD

By adding, we get XP+PY=21BC+21AD

∴XY=21(BC+AD)

Answer 2

Hy

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Here we will prove that the line segment joining the midpoints of the nonparallel sides of a trapezium is half the sum of the lengths of the parallel sides and is also parallel to them. ... U and V are the midpoints of QR and PS respectively. To prove: (i) UV ∥ RS. (ii) UV = 12(PQ + RS).

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