Basic Geometry Theorem

Tag:
BASIC PROPORTIONALITY THEOREM

If a line is drawn parallel to one side of a triangle, the other two sides of the triangle are divided proportionally.


Thus, in Fig. DE || BC, According to the above result


AD/DB = AE /EC

We can easily verify this by measuring AD, DB, AE and EC.
We state the converse of the above result as follows :

If a line divides any two sides of a triangle in the same ratio, the line is parallel
to third side of the triangle.

Example:

if BC is parallel to DE and AB=2, BD=3, AC=4 then what what is AE

Ans: AB/BD=AC/CE =>2/3=4/CE
=>CE =6 and AE = 10

try A tough one


In the above figure angle A is 30 degree and length of BC is 3 cm, then find out the length of

a. DE
b. AB, BD
c. AC, CE
__________________

n/a

Drisya's picture
Offline
Last seen: 9 years 2 weeks ago
Joined: 2008-10-09
pretty fine theorem!!!!!!!

That is pretty clear theorem. But I 'm cofused about the question posted. I could not come up with a solution. Could someone help me out with this.

Thanks in advance.

rahulv1010's picture
Offline
Last seen: 7 years 18 weeks ago
Joined: 2010-06-21
the theorem was gud but the

the theorem was gud but the question was incomplete or the data given was insufficent u cn only calculate the length of ab and ac...

__________________

n/a

__________________

n/a

Incognito
Incognito's picture
Not exactly,all the lengths

Not exactly,
all the lengths can be calcualted as this is a right angle triangle

rahulv1010's picture
Offline
Last seen: 7 years 18 weeks ago
Joined: 2010-06-21
nah ....nt exactly all the

nah ....nt exactly all the lengths .... the lengths BD nd CE have to be confined with certain constraints .....as no such constraints is given the lengths BD CE nd DE cn acquire any possible length keeping the angle 30 degree nd the sides AB nd AC fixed accordingly ..... give it a try u wont be able to find any particular nd fixed length for BD CE nd DE there cn be infinite possibilities nd infinite probable lengths for the 3 sides even though there are right angled triangles in above fig.

__________________

n/a

__________________

n/a

abhik_07's picture
Offline
Last seen: 9 years 25 weeks ago
Joined: 2008-05-29
Answers are

My answers are
AB = 3 sqrt(3)
AD = 6 sqrt(3)
AC = 6
AE = 12
DE = 6

 

ISHANI (not verified)
Incognito's picture
MY ANSWERS

MY ANSWERS ARE
DE=3
AB = 3 sqrt(3)
AD = 6 sqrt(3)
AC=3
CE=3

Post new comment

The content of this field is kept private and will not be shown publicly.

Sponsered Links

All Rights Reserved. Copyright 2006-10 CAT4MBA.com.