Question

Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Using the above result, do the following:
In Fig. 7, DE||BC and BD = CE. Prove that ΔABC is an isosceles triangle.

Answer 1

As, DE||BC and AD/DB = AE/CE
we have BD = CE
AD/BD = AE/BD
AD=AE
So, we get
AD+DB=AE+EC
AB= AC
Hence,.Δ ABC is an isosceles triangle...
(Proved)

Answer 2

Given :A triangle ABC in which DE //BC , and intersects AB in D and AC in E .

To Prove: $\frac{AD}{DB}=\frac{AE}{EC}$

Construction: Join BE,CD and draw EF perpendicular to BA and DG perpendicular to CA.

Proof:

Since ,EF is perpendicular to AB .

Therefore , EF is the height of ∆ADE and ∆DBE.

Now, Area (∆ADE)=1/2(Base×height)

= 1/2(AD×EF)

and, Area (∆DBE) = 1/2(DB×EF)

$\frac{Area(triangle ADE)}{Area(triangleDBE)}=\frac{\frac{1}{2}(AD\cdot EF)}{\frac{1}{2}(DB\cdot EF)}=\frac{AD}{DB}$---(1)

Similarly, We have

$\frac{Area(triangleADE)}{Area(triangleDEC)}=\frac{\frac{1}{2}(AE\cdot DG)}{{\frac{1}{2}(EC\cdot DG)}=\frac{AE}{EC}$---(2)

But ,∆DBE and ∆DEC are on the same base and between the same parallels.

Therefore,

Area(∆DBE) = Area(∆DEC)

$\implies \frac{1}{Area(triangleDBE)}=\frac{1}{Area(triangleDEC)}$

/* Taking reciprocals of both sides*/

$\implies \frac{Area(triangleADE)}{Area(triangleDBE)}=\frac{Area(triangleADE)}{Area(triangleDEC)}$

/* Multiplying both sides by Area (∆ADE)*/

$\frac{AD}{DB}=\frac{AE}{EC}$/* from(1)&(2)*/

___________________

ii) In ∆ABC , DE//BC and BD=CE

$\frac{AD}{DB}=\frac{AE}{EC}$
/* By above theorem*/

$AD=AE$

\* GivenBD=CE*\

Now, AB = AD+DB

= AE+CE

= AC

Therefore,

AB = AC

then ABC is an isosceles triangle.

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