## BdMO National Secondary/Higher Secondary 2018/2

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### BdMO National Secondary/Higher Secondary 2018/2

$AB$ is a diameter of a circle and $AD$ & $BC$ are two tangents of that circle.$AC$ & $BD$ intersect on a point of the circle.$AD=a$ & $BC=b$.If $a\neq b$ then $AB=?$

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

### Re: BdMO National Higher Secondary 2018/2 Capture.PNG (20.03 KiB) Viewed 1024 times
Let $AC$ and $BD$ meets at $E$.
$AD=a,BC=b$ where, $a<b$.
Easy to prove: $\angle CAD=\angle ABE, \angle DBC=\angle BAE$
We know, $\angle BAD=\angle BAE+ \angle CAD=90^{\circ}$
$\Rightarrow \angle BAE+\angle ABE=90^{\circ}$
Again, $\angle ABE+\angle ADB=90^{\circ}$
So, $\angle BAE=\angle ADB$

Accordingly we can prove: $\angle ABD=\angle ACB$

Between $\triangle ABD$ and $\triangle ABC$,
$\angle BAD=\angle ABC$,$\angle ADB=\angle CAB$
So, two triangles are similar.

Now, $\frac {a}{AB}=\frac{AB}{b}$
$\Rightarrow AB^2=ab$
$\Rightarrow AB=\sqrt {ab}$

[Actually I messed all similar triangles and wasted time at exam hall for this easy one ]