Orthocentre of triangle in cyclic qudrilateral

Question

From vertex B of parallelogram $ABCD$ heights $BK$ and $BH$ are drawn on sides $AD$ and $DC$ respectively. It is known that $KH = a$ and $BD = b$. Find the distance from $B$ to the intersection points of the heights of triangle BKH. Hint: use some translational transformation.

Here's what I've done so far -- the regions $CHB$ and $KAB$ are irrelevant to the question, and we need to only consider the quadrilateral $DHBK$, which is cyclic, because $\angle DHB + \angle DKB = \pi$. Further, we see that since $\angle DHB =\angle DKB = \pi/2$, $BD$ is the diameter of the circumcircle of $DHBK$. I have no clue on how to proceed to find the distance from $B$ to $BHK$'s orthocentre.

Answer 1

Lemma (well-known). Let $ABC$ be a triangle with orthocenter $H$ and circumradius $R$. Then $AH=2R\cos A$.

Proof. Let the antipode of $B$ in the circumcircle be $B'$. We claim that $AB'CH$ is a parallelogram. Indeed, $AH\perp BC\perp B'C$ and $B'A\perp AB\perp CH$, and the claim follows.

Thus $AH=B'C=BB'\cos A=2R\cos A$.

---

Let $H'$ be the orthocenter. Then from our lemma $BH'=b\cos\angle HBK$.

The sine rule in $\triangle BHK$ tells us that $\sin\angle HBK=\frac{a}{b}$. Now

$$\cos\angle HBK=\sqrt{1-\sin^2\angle HBK}=\sqrt{1-\frac{a^2}{b^2}},$$ so $BH'=\sqrt{b^2-a^2}$.

Answer 2

Let's call $J$ the orthocenter of $BHK$. We can use the known theorem (http://demonstrations.wolfram.com/TheDistanceToTheOrthocenterFromAVertex/):

$$(BJ)^2+(KH)^2=(2R)^2 \quad (1)$$

where $R$ is the radius of the circumcircle of $BHK$.

But the circumcircle of $BHK$ is also circumcircle of $DHBK$ and then $2R=b$.

Backing to $(1)$ we get:

$$(BJ)^2+a^2=b^2 \Rightarrow BJ=\sqrt{b^2-a^2}$$