In this Blog we will talk about a fascinating geometry lemma which can help you to solve some of the hardest and finest geometry problems around the globe, so without any delay let's look up to the lemma.
: In a 
, a line from 
intersects 
at 
and a line from 
intersects 
at 
and 
be the intersection point of 
and 
, then 
and 
(SEE THE FIGURE FOR REFERENCES)
Let's work through the proof:
construction! , yes , so we construct 
. Then we observe 
and 
.
Then we have 
and 
.
Then also we have 
. So we have 
. Also, .
So simplifying we get 
so we get 
. Similarly, we can get the other ratio. 
Now Let's Look up to some applications of this Lemma.
If 
is the angle bisector of 
in , s.t. 
and 
is it's Incenter, show that: 
.png)
so, we do some mark-ups, denote 
, 
and 
, the angle bisector of 
hits , 
at and similarly the angle bisector of 
hits at
Now we use the angle bisector theorem to get: 
, now from the lotus ratio lemma, we get:
.
hence our problem is done!
As shown in the figure, triangle is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangle are as indicated. Find the area of triangle . (AIME 
)
Firstly denote point intersection of line from 
and 
be using the lotus ratio lemma we notice: 
, from base division theorem , 
and 
. So we have 
, so again from base division theorem ![\frac{[ACF]}{[ABF]}=\frac{1}{2}](https://sumonmath.com/wp-content/ql-cache/quicklatex.com-f6ce33add41821ba8333700e8bbb2304_l3.png "Rendered by QuickLaTeX.com")
, so ![[ACF]=210](https://sumonmath.com/wp-content/ql-cache/quicklatex.com-7bb1d871f379a7230dfcd1a1852c193c_l3.png "Rendered by QuickLaTeX.com")
hence ![[ABC]=210+105=\boxed{315}](https://sumonmath.com/wp-content/ql-cache/quicklatex.com-a04fd5b165a2df0f0fb1c320d0367b86_l3.png "Rendered by QuickLaTeX.com")
,
interestingly there was no use of giving area 
.(Which actually shows how powerful the lemma is!)
(Menelaus' Theorem)
If line 
intersecting on 
, where is on 
is on the extension of , and 
on the intersection of and , then
.
Now, you can try to give a rigid proof of this problem using lotus ratio lemma , for further discussions. Join us at SuMON Slow Math Program to know more interesting lemmas and theorems like these while exploring problems from the problem sets, solve more exciting problems like this every fortnight, which require only logical thinking rather than conceptual prerequisites making math fun! These kind of problems are not available in wide range of math text books. We also provide detailed solution to each problem for every fortnight which will teach you a lot of fun lemma's and ideas in mathematics.
Happy mathematics!
