Dharmath Vader: Solution: Cyclic quadrilateral

Thursday, May 6, 2010

$AB=AD=x, BC=y$ and

$CD=x+z$ with

$y < z$ .

Let

$\theta = \angle ADC$ so

$\angle ABC = 180^o-\theta$ . It's easy to see that because

$CD > BC$ then

$\angle ABC > \angle ADC$ thus

$0 < \theta < 90^o$

Now

$AC^2 = x^2 + y^2 + 2xy \cos \theta = x^2 + (x+z)^2 - 2x(x+z)\cos \theta$ , simplify it to get:

$\cos \theta = \frac{(x+z)^2-y^2}{2x(x+y+z)} = \frac{x+z-y}{x}$

But since

$\theta$ is an acute angle,

$\theta < 60^o \iff \cos \theta > 1/2$
So we need to show

$x+z-y > x$ which is true because

$z > y$

Second Solution

Let

$R$ be a point on

$CD$ such that

$CB = CR$ (see the picture).

Since

$AB=AD$ , then

$CA$ is an internal angle bisector, and thus

$\triangle ABC$ are congruent to

$\triangle ARC$ . That means

$RD = CD - CR = CD - CB > AB = AD$ .
Also,

$AD = AB = AR$ .
So

$\triangle ARD$ is an isosceles where

$AD = AR$ and

$RD > AD$ , which means that

$\angle ADR < 60^o$ and thus

$\angle ABC > 120^o$

Dharmath Vader