write and prove midpoint theorem Converse of midpoint theorem and intercept theorem with figure.....
Answers
Converse of mid-point theorem: it states that in a triangle line drawn from the mid-point of the one side of triangle, parallel to the other side intersect the third side at its mid-point. ... therefore by the AAS congruency triangles are congruent. thus AE = EC. i.e.
HOLA MATE
ANSWER:-
Let E and D be the midpoints of the sides AC and AB. Then the line DE is said to be parallel to the sides BC, whereas the side DE is half of the side BC; i.e.
DE is parallel to BC
DE∥BC
DE = (1/2 * BC).
Now consider the below figure,
Mid- Point Theorem
Construction- Extend the line segment DE and produce it to F such that, EF=DE.
In the triangle, ADE, and also the triangle CFE
EC= AE —– (given)
∠CEF = ∠AED {vertically opposite angles}
EF = DE { by construction}
hence,
△ CFE ≅ △ ADE {by SAS}
Therefore,
∠CFE = ∠ADE {by c.p.c.t.}
∠FCE= ∠DAE {by c.p.c.t.}
and CF = AD {by c.p.c.t.}
The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.
In a similar way, ∠FCE and ∠DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.
Therefore, CF∥AB
So, CF∥BD
and CF = BD {since BD = AD, it is proved that CF = AD}
Thus, BDFC forms a parallelogram.
By the use of properties of a parallelogram, we can write
BC∥DF
and BC = DF
BC∥DE
and DE = (1/2 * BC). Hence Proved.
HOPE IT HELPS.
