Question

In ∆ABC, seg DE || Side BC. If 2A (∆ADE) = A( DBCE), find AB : AD and show that
BC = √3 X DE​

Answer 1

Answer:

i hope the answer is corrrect

Answer 2

Given:

In ΔABC, segment DE ║ side BC

2A(ΔADE) = A(DBCE)

To Find:

BC = √3×DE

Solution:

Since DE is parallel to BC

In ΔADE and ΔABC,

∠A is common

The corresponding angles of a triangle are equal

∴ ∠ADE = ∠ABC

So, ΔADE is similar to ΔABC by AA similarity criteria.

∴ $\frac{AD}{AB} = \frac{DE}{BC} = \frac{AE}{AC}$  ..(i)

Then,

$\frac{A(triangle ADE)}{A(triangle ABC)}$ = $\frac{AD^{2} }{AB^{2} } = \frac{DE^{2} }{BC^{2} } = \frac{AE^{2} }{AC^{2} }$  ..(ii)

Also, 2A(ΔADE) = A(DBCE) ..(iii)     [given]

Since, A(ΔABC) = A(ΔADE) + A(DBCE)

          A(ΔABC) = A(ΔADE) + 2A(ΔADE)

∴ A(ΔABC) = 3A(ΔADE)

$\frac{A(triangle ADE)}{A(triangle ABC)}$ = $\frac{1}{3}$ ..(iv)

Using (ii) and (iii), we get

$\frac{AD^{2} }{AB^{2} } = \frac{DE^{2} }{BC^{2} } = \frac{1}{3}$

∴ $\frac{AD^{2} }{BD^{2} } = \frac{1}{3}$  and $\frac{DE^{2} }{BC^{2} } = \frac{1}{3}$

$\frac{AD}{BD} = \frac{1}{\sqrt{3} }$   ..(v)  and $\frac{DE}{BC} = \frac{1}{\sqrt{3} }$ .(vi)

Now, by applying inverted in (v)

We get,

$\frac{AB}{BD} = \sqrt{3}$ and

Using equation (vi), we get

BC = $\sqrt{3}$ × DE

Hence proved.