Question

plz answer this question .
(Class 9th )​

Answer 1

Answer:

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.”

Answer 2

⚘ Midpoint theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.

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Given

• A ∆ ABC in which D and E are the midpoints of AB and AC respectively

• DE is joined.

To prove

$\sf{ \to DE \: || BC \: and \: DE \: =\frac{1}{2}BC}$

Construction

»Draw CF || to BA, meeting DE produced in F

Proof

In ∆ AED and ∆ CEF , we have

$\sf{⦿ \: \angle AED= \angle CEF \: \: \: [vertically \: opposite \: angles]}$

$\sf{⦿AE=CE \: \: \: \: [\because E \: is \: midpoint \: of \: AC]}$

$\sf{⦿\angle DAE=\angle FCE \: \: \: \: [ alternate \: interior \: angles]}$

$\mathtt{\therefore\red{∆ AED≅∆ CEF (ASA \:criterion)}}$

And so, AD = CF and DE = EF (c.p.c.t)

$\sf{But, AD = BD \: \: \: \: [\because D \: is \: the \: midpoint \: of \: AB]}$

$\sf{and \: BD \: || \: CF \: \: \: \: (by \: construction)}$

$\sf{\therefore BD= CF \: and \: BD || \: CF}$

$\mathtt{\implies \orange{BCFD \: is \: a \: parallelogram}}$

$\sf{\implies DF \: ||BC \: and \: DF= BC}$

$\sf{\implies \green{DE|| \: BC \: and \: DE =\frac{1}{2}DF =\frac{1}{2}BC}}$

$\mathtt{➣ \pink{Hence \: DE \: || \: BC \: and \: DE =\frac{1}{2}BC}}$

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