Please give me Class 9 ICSE solutions of Mid point theorem..
Answers
Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined.
Given: AD = DB and AE = EC.
To Prove: DE ∥∥ BC and DE = 1212 BC.
Construction: Extend line segment DE to F such that DE = EF.
Proof: In △△ ADE and △△ CFE
AE = EC (given)
∠∠AED = ∠∠CEF (vertically opposite angles)
DE = EF (construction)
hence
△△ ADE ≅≅ △△ CFE (by SAS)
Therefore,
∠∠ADE = ∠∠CFE (by c.p.c.t.)
∠∠DAE = ∠∠FCE (by c.p.c.t.)
and AD = CF (by c.p.c.t.)
The angles ∠∠ADE and ∠∠CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.
Similarly, ∠∠DAE and ∠∠FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.
Therefore, AB ∥∥ CF
So, BD ∥∥ CF
and BD = CF (since AD = BD and it is proved above that AD = CF)
Thus, BDFC is a parallelogram.
By the properties of parallelogram, we have
DF ∥∥ BC
and DF = BC
DE ∥∥ BC
and DE = 1212BC (DE = EF by construction)
Hence proved.
I HOPE THIS WILL HELP YOU..
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, the midpoint theorem is Proved.