Let $ABCD$ be a convex quadrilateral and $M$ be the midpoint of the segment $BC$ such that $$∠AMD = 90^\circ,\ ∠ADM = 15^\circ,$$ and $$AD = AB + CD.$$ Find $∠BAD$.
I try to find an additional construction but I don't know. Can you help me with one idea?
Let us define an point $X$ on $AD$. Construct a circle,$Σ$ of radius $AX$, taking point $A$ as a center.
Construct another circle,$Ω$ of radius $(AD-AX)$, taking $D$ as a center.
Point $M$ must lie on the tangent to $Σ$ and $Ω$ (whose points of contact on $Σ$ and $Ω$ define the points $B$ and $C$ respectively).
Also, $M$ becomes the midpoint of $BC$ as tangents drawn from an external point to a circle are equal. $(MB=MX=MC)$.
Join $MX$ (which becomes a perpendicular to $AD$)
This gives us an imaginary contruction.
Tangents draw to a circle are perpendicular to the radius at the point of contact. So, angles $MBA$ and $MCD$ are right angles.
Also, tangents drawn to a circle from a given point are equally inclined to the line joining the center to the point. So, angles $BMA$ and $XMA$ are equal and are both $15°$ $(90°-MAD=90°-75°)$.
Thus, in $ΔBAM$, angle $BAM=75°$.
And finally, Angle $BAD=150°$. Ans.
