Prove parallel

[asif e elahi]

Let $ABCD$ be an isosceles trapezium so that $AD\parallel BC$ and $AB=CD$. Let $\omega$ be the incircle of $\triangle BCD$ and $\Gamma$ be any circles that touches $\overline{AB}$ and $\overline{AC}$ and intersects $BC$ at $U$ and $V$. Let $DU$ and $DV$ meet $\omega$ at $J$ and $K$ respectively. Prove that $JK\parallel BC$.

[rah4927]

Hint : Forget about the trapezoid. Think about how the second circle looks like in $\triangle DBC$.Apply POP .

[Raiyan Jamil]

Solution:

We reflect the whole figure except the triangle $ABC$ under the perpendicular bisector of $BC$. Let the Point $X'$ represent the reflection of a point $X$. Now, we can easily prove by various calculations that there exist a circle tangent to both $AB$ and $AC$ and passes through the points $U'$ and $V'$. So, taking $A$ as the homothetic center we get that $J'K'||U'V'$ or $JK||UV$ or $JK||BC$ as desired.