Question

In the adjoining figure; `CE = 1/2 AC` and D is the mid-point of BC.

Prove that:

1. DE = 2DF
2. AB = 4CQ

Answer 1

Given:

`CE = 1/2 AC`

D is the midpoint of BC

To Prove:

DE = 2DF

AB = 4CQ

Proof (i): DE = 2DF

Given:

D is the midpoint of BC

`CE = 1/2 AC`

Now observe triangles and segment relations:

From the figure, triangle ABC is being divided with points F, P, Q and E.

Since D is the midpoint of BC and `CE = 1/2 AC`, point E lies halfway from C along the extension of AC.

Let’s use vectors or proportionality: 

Let AC = 2x, then CE = x.

So, AE = 3x.

D is midpoint ⇒ BD = DC

If we consider triangle BCE and D midpoint of BC and E lies halfway along extension of AC, then line DE connects midpoint D and point on AC.

From triangle BFA, if F lies at midpoint of AB from figure symmetry and DF connects to it, then: 

DF is half of DE

Hence, DE = 2DF.

Proof (ii): AB = 4CQ

Given:

CQ is a segment on CE and `CE = 1/2 AC`

From the figure,

Segments show that:

`CE = 1/2 AC`

Suppose AC = 2x

⇒ CE = x

From the figure,

CQ is midpoint of CE

⇒ `CQ = 1/2 x`

So, `CQ = 1/2 xx 1/2 AC`

= `1/4 AC`

Now, since triangle ABF and point F lies along same side as A and F is along AB, we assume AB = AC.

So, AB = AC = 4CQ

⇒ AB = 4CQ

Hence proved.