MID POINT THEOREM: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
GIVEN A in which D and E are the mid-points of sides AB and AC respectively. DE is joined. TO PROVE CONSTRUCTION Produce the line segment DE to F, such that DE = EF. Join FC. PROOF In , we have AE = CE [ E is the mid-point of AC] [ Vertically opposite angles] and, DE = EF [By construction] So, [By SAS criterion of congruence] AD = CF [C.P.C.T.] .....(i) and, [C.P.C.T.] .....(ii) As it is given that D is the mid-point of AB [From (i) we know that AD = CF] .....(iii) From Eq 2 We know that Now, But they are alternate interior angle when AB and CF are straight lines and DF is the transversal As they are equal so AB || FC , or BD || CF [ When whole are parallel the parts are parallel] Hence BD = CF and BD || CF BCFD is a IIgm [ As one opposite pair of sides is parallel and equal] DF || BC and DF = BC [ Opposite sides of a are equal and parallel] But, E is the midpoint of DF [ By construction DE = EF ] Hence proved. | |
Illustration:ABCD is a rhombus, EABF is a straight line such that EA = AB = BF Prove that ED and CF when produced meet at right angle. | |
Proof: ABCD is a rhombus and diagonals of a rhombus bisect at right angle In A is the mid point of EB [ Given EA = AB] O is the mid point of BD [ Diagonals bisect] Now as O and A are mid points of BD and EB . By mid point theorem we can say that ED || AO EG || AC Similarly In B is the mid point of AF [ Given AB = BF] O is the mid point of AC [ Diagonals bisect] Now as O and B are mid points of AC and AF . By mid point theorem we can say that CF || BO FG || BD AS BD || FG and AC is the transversal .................(1) [Corresponding angle] Similarly EG || AC and GF is the transversal .................(2) [Corresponding angle] From (1) and (2) We get , Hence Proved |
ABCD is a parallelogram in which P,Q,R,S are mid-points of the sides AB, BC, CD and DA respectively. AC is a diagonal. Then which of the following is true? | |||
Right Option : D | |||
View Explanation |
P, Q, R are respectively, the mid-points of sides BC, CA and AB of a triangle ABC. PR and BQ meet at X. CR and PQ meet at Y. Then which of the following is true? | |||
Right Option : D | |||
View Explanation |
ABCD is a parallelogram in which P,Q,R,S are mid-points of the sides AB, BC, CD and DA respectively. AC is a diagonal. Then which of following is true? | |||
Right Option : C | |||
View Explanation |
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