Show that the triangle $ADC$ is isosceles

Question

Let $K$ be a circle with center $M$ and $L$ be a circle that passes through $M$ and intersects $K$ in two different points $A$ and $B$ and let $g$ be a line that goes through $B$ but not through $A$. Let $C$ be the second point of $g$ on $L$ and let $D$ be the second point of $g$ on $K$ with $C=B$ or $D=B$ if $g$ is the tangent through $B$ on the circle $K$ or $L$. Show that the triangle $ADC$ is isosceles.

For that do we have to show that that the length of each side of triangle is equal? Or how can we show that?

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EDIT:

We get the following, or not?

Answer 1

Hint:

Let $ \angle ADC = \alpha$.
What is $ \angle AMB$ in terms of $ \alpha$?
What is $ \angle ACB$ in terms of $ \alpha$?
What is $ \angle DAC$ in terms of $ \alpha$?

Answer 2

Let's look at the two tangent cases also : when $g$ is tangent to $K$ (call it $g_1$) and when $g$ is tangent to $L$ (then call it $g_2$). Both can be shown on same diagram.

Let $\angle ADB = \alpha$. So $\angle AMB = 2\alpha$ and $\angle ACB=180-2\alpha$.

Now as $g_2$ is tangent to red circle, $\angle ABD = \angle ACB=180-2\alpha$ by alternate segment theorem. Similarly as $g_1$ is tangent to black circle, $\angle ABC= \angle ADB=\alpha$.

The third angles can be calculated. $\angle DAB = \alpha = \angle BAC$.

We have found both $\triangle ABD$ and $\triangle ACB$ are isosceles and similar.

Note :

According to question statement, in case of $g_1$, $B=C$ , $\triangle ADB$ in diagram is $\triangle ADC$. While in case of $g_2$, $B=D$, $\triangle ABC$ in diagram is $\triangle ADC$. So we have shown in both cases $\triangle ADC$ (according to question) is isosceles.

Answer 3

Here is another possible construct -

Please note the proof is very similar. Quadrilateral $AMBC$ is cyclic so

$\angle ACD = 180^0 - \angle AMB$

$\angle ADB = 180^0 - \frac{1}{2} \angle AMB \implies \angle ADC = \frac{1}{2} \angle AMB$

Given $\angle ACD$ and $\angle ADC \,$ in $\triangle ADC, \angle CAD = \frac{1}{2} \angle AMB$.

So $\triangle ADC$ is isosceles.