Question

please send Ch 10 mid point theorem solutions of ML aggarwal icse maths book for class 9

Answer 1

HEYA FRIENDS, HERE IS YOUR ANSWER :


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 ANSWER WILL HELP YOU

Answer 2

MidPoint Theorem Statement

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

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.