## Dhaka Secondary 2009/8

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
Forum rules
Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### Dhaka Secondary 2009/8

For triangle $ABC$ $\angle C$ is $90^{\circ}$. $\angle BAC$ is $30^{\circ}$ & $AB$ is $1$cm. $D$ is a point within $ABC$ so that angle $\angle BDC$ is $90^{\circ}$ & $\angle ACD = \angle DBA$. $AB$ & $CD$ meets at $E$. Find $AE$.

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Dhaka Secondary 2009/8

ans is $\frac{1}{2}$. i just used similiar triangle properties and general trigonometric means.
বড় ভালবাসি তোমায়,মা

Moon
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Contact:

### Re: Dhaka Secondary 2009/8

^By outline I did not mean that you should just post what you used to solved the problems (it is more or less apparent that some Euclidean geometry and trig can solve almost any problem at this level). You better post what you used to solve what.

It is also best for everyone if you post complete solution. Remember, in National we need to write complete solutions.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

leonardo shawon
Posts: 169
Joined: Sat Jan 01, 2011 4:59 pm
Location: Dhaka

### Re: Dhaka Secondary 2009/8

tahmid, full sollution or at least a little hint.. Please
Ibtehaz Shawon
BRAC University.

long way to go .....

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Dhaka Secondary 2009/8

sorry i'll try from next time and post almost full but not formal solns
বড় ভালবাসি তোমায়,মা

Shapnil
Posts: 10
Joined: Mon Dec 19, 2011 2:05 pm

### Re: Dhaka Secondary 2009/8

Plz give the full solution or a hint

Ridwan Abrar
Posts: 13
Joined: Mon Aug 05, 2013 8:01 pm

### Re: Dhaka Secondary 2009/8

in triangle ABC, cos 30=AC/1
ANGLE DCB+ ANGLE DBC=90
ANGLE ACD+ANGLE DCB=90
GIVEN, ANGLE ACD=ANGLE DBA
SO, ANGLE ACD=ANGLE DBA
ANGLE DBC=ANGLE DBA=60/2=30
sin(120)/AC=sin(30)/AE
AE=0.5

mdhasib
Posts: 10
Joined: Fri Jul 22, 2016 11:43 am

### Re: Dhaka Secondary 2009/8

We can easily show that triangle FCD and triangle EDB are similar . So FD/DE = CD/BD = FC/BE . Again we can see that triangle FCD and triangle BCD are similar . Again FD/CD = CD/BD = FC/BC .If we merge the conclusions we get FD/CD = FC/BC = FD/DE = CD/BD = FC/BE. We have FC/BC = FC/BE hence BC=BE. Now we can apply trigonometry and get the value of BE then i think we all know what to do

[Q.E.D]

SYED ASHFAQ TASIN
Posts: 29
Joined: Thu Jun 02, 2016 6:14 pm