Trapezoid proof. If $BD$ is isosceles trapezoid's diagonal, $AB$ and $CD$ are sides, $BC$ and $AD$ are bases, prove that $BD^2=AB^2+AD \cdot BC$.
 A: Hint: assume WLOG that $BC$ is the larger base. Let $H$ be the foot of the perpendicular from B onto line $AC$. Apply Pythagora's theorem to the two right triangles $\triangle BHA$ and $\triangle BHD$ then eliminate $BH$ between the equations:
$$
BA^2 = BH^2 + \left(\frac{BC-AD}{2}\right)^2 \\
BD^2 = BH^2 + \left(\frac{BC+AD}{2}\right)^2
$$
A: A solution relying on pictures and geometric constructions.

On the extension of line $BC$, draw point $F$ so that $CF = AD$ and $C$ is between $B$ and $F$ like on the picture. Since lines $BC$ and $AD$ are parallel and $F$ lies on line $BC$, the segments $AD$ and $CF$ are parallel and equal in length. Therefore quad $CFDA$ is a parallelogram and so $AC$ and $DF$ are parallel and because $ABCD$ is isosceles trapezoid, $BD = AC = DF$. 
Furthermore, on the line $DC$ choose points $E$ and $K$ as shown on the picture so that $BD = DE = DK$, i.e. $D$ is the midpoint of $EK$. Now, observe that 
$BD = DF = DE = DK$, so the points $B, E, F, K$ lie on the circle, we denote by $k_D$, centered at point $D$ and of radius $BD$. 
Apply the power of point relation to the chords $BF$ and $EK$ intersecting at point $C$ and obtain the identity
$$BC \cdot CF = EC \cdot CK$$  But since $CF = AD$ as well as $$CK = CD + DK = AB + BD \,\, \text{ and } \,\, EC = DE - DC = BD - AB$$ we substitute these identities in the power of point relation 
$$BC \cdot AD =  BC \cdot CF = EC \cdot CK = (BD - AB) (BD + AB) = BD^2 - AB^2$$ leading to the desired identity
$$BD^2 =  AB^2 + BC \cdot AD$$
