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### Dhaka Secondary 2011/6

Posted: **Fri Jan 28, 2011 10:04 pm**

by **BdMO**

The diagonal $AB$ in quadrilateral $ADBC$ bisects the angle $CBD$. $DB$ and $DA$ are tangent to the circumcircle of triangle $ABC$ at points $A$ and $B$. The perimeter of triangle $ABC$ is $20$ and perimeter of triangle $ABD$ is $12$. Find the length of $BD$.

### Re: Dhaka Secondary 2011/6

Posted: **Tue Feb 01, 2011 1:48 am**

by **Avik Roy**

I'm a bit sad seeing that people are not trying this problem. It's one of my favourite. As a divisional problem, it takes good count of your geometric senses and algebraic capabilities

### Re: Dhaka Secondary 2011/6

Posted: **Tue Feb 01, 2011 11:12 am**

by **Tahsin24**

I've solved it n my ans is (36/11)......I'm sorry that I cant post a full solution because I cant use LATEX yet. I'll try to learn it soon. My procedure was to show tringle ABC and triangle ABD are both isosceles and similar to each other.

### Re: Dhaka Secondary 2011/6

Posted: **Tue Feb 01, 2011 6:46 pm**

by **Tahmid Hasan**

i think it's $\frac{36}{11}$

i did something with the ecantor brittangshostho thingyy.

### Re: Dhaka Secondary 2011/6

Posted: **Tue Feb 01, 2011 11:36 pm**

by **Moon**

Tahmid: Please post your solution!

### Re: Dhaka Secondary 2011/6

Posted: **Tue Feb 01, 2011 11:38 pm**

by **Tahmid Hasan**

for that i'll have to draw the pic but i don't know how to post that here.

so if any1 could give a pic i' gladly give the soln

### Re: Dhaka Secondary 2011/6

Posted: **Tue Feb 01, 2011 11:55 pm**

by **Moon**

Posting diagrams is easy. Just create a diagram and attach it with the post

(There is an "upload attachment" option below post reply editor...however you must use the "full editor" to get this option.)

### Re: Dhaka Secondary 2011/6

Posted: **Sat Feb 05, 2011 3:55 pm**

by **photon**

i find ABD isosceles.how did you get ABC as isosceles?

### Re: Dhaka Secondary 2011/6

Posted: **Sat Feb 05, 2011 3:55 pm**

by **photon**

how 2 get similarity???

### Re: Dhaka Secondary 2011/6

Posted: **Mon Feb 07, 2011 12:18 am**

by **Tahsin24**

$ABD$ and $ABC$ are both isosceles. $\angle BAD=\angle ACB$ (ekanttor brittangshostho kon),

so $\angle ACB=\angle ABD=\angle BAD=\angle ABC$

so in triangle $ABC$ and $ABD$

$\angle ABC=\angle ABD$

$\angle ACB=\angle BAD$

$\angle BAC=\angle ADB$