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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 |
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Answer» 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) |
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