Question

Hello!!!!

So the question is in the attachment.....
Question (c)

Hoping the best solution:)

Answer 1

$\huge{\pink{\underline{\bold{Answer}}}}$

In ∆ ABC,

D is the mid point of AB and DF ll BC

F is the mid point of AC .....(a)

How it is by mid point theorem (converse)

Statement

$The \:line \:drawn \:through \:the\: mid \\point \:of \:one \:side \:of \:triangle \\, parallel \:to \:another \:side \:bisects \\the\: third\: line.\\ ( hence \:proved)$

Now, F and E are mid points of AC and BC respectively.

EF ll AB .....(b)

Now DF ll BC

Now DF ll BE ......(c)

EF ll AB .....[from b]

$\:\red{\boxed{EF \parallel DB}}$ .....(d)

From (c) and (d)

$\red{\boxed{DBEF \:is \:a \:parallelogram}}$

(2) F is mid point of AC

AC = 2 AF

=> 2 × 2.6 cm

$\pink{\boxed{\bold{\implies{AC= 5.2 \:cm}}}}$

Answer 2

Answer:

Given :

In ∆ ABC,

D is the mid point of AB and DF ll BC

F is the mid point of AC [ Converse of Midpoint Theorem ]

Now, F and E are mid points of AC and BC respectively.

EF ll AB -----(i)

DF ll BC

Now DF ll BE ------(ii)

EF ll DB ------(iii)

So , DBEF is a parallelogram from (ii) and (iii) .