This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
ABCD is a trapezium with AB parallel to CD and midpoints of AC and BD be P and Q repectively........ CONSTRUCTION - Draw a line from D passing through Pand meeting AB in E............PROOF: In triangle APE and triangle CPD angle PAE = angle PCD ( alternate angles ) Angle PEA = angle PDC ( alternate angles ) AP = PC ( P is the midpoint of AC ) triangle APE is congruent to triangle CPD PE = PD ( CPCT ) In triangle BED P is the midpoint of DE ( proved above ) Q is the midpoint of BD ( given ) Therefore PQ is parallel to BE ( midpoint theorem ) SO PQ is parallel to AB AB parralel to CD ( given ) SO PQ is parallel to CD and AB And one more step is...Join DE and produce to meet AB at G In triangle AEG and CEG <1 = <2 (Alternate interior angles) AE = EC (E is the mid point of AC) <4 = <3 ( Vertically opposite angles) Triangle AEG is congruent to triangle CED (by ASA congruence rule) DE = EG ( CPCT) AG = CD ( CPCT) In triangle BGD E is the mid point of DG F is the mid point of BD By mid point theorem, EF II BG EF II AB AB II CD Hence, AB II EF II CD EF = 1/2 BG EF = 1/2 ( AB - AG) EF = 1/2 (AB - CD) HOPE THIS HELPS :)
3 5 3