How can pove a generalization of Brahmagupta theorem

Question

In the figure:

If $\frac{MQ}{BA}=\frac{NP}{BC}$ and $\angle MBA = \angle CBN=\frac{\pi}{2}$ And $D, E$ be the midpoint of $MN, PQ$ respectively, then show that $DE \perp AC$ If $P=Q=B$ this is a generalization of Brahmagupta theorem (Figure 2)

If $P=Q=B$ and $M, B, C$ are colliear, and $N, B, A$ are collinear; This case is Brahamagupta theorem

Answer 1

Let $\vec{BA}=\vec{a}$, $\vec{BC}=\vec{c}$, $\vec{QM}=\vec{m}$ and $\vec{PN}=\vec{n}$.

Thus, $$\vec{ED}=\frac{1}{2}(\vec{m}+\vec{n})$$ and we need to prove that $$(\vec{c}-\vec{a})(\vec{m}+\vec{n})=0$$ or $$\vec{c}\vec{m}=\vec{a}\vec{n},$$ which is obvious because $$\frac{|\vec{m}|}{|\vec{a}|}=\frac{MQ}{BA}=\frac{NP}{BC}=\frac{|\vec{n}|}{|\vec{c}|}$$ and $$\measuredangle MBC=\measuredangle NBA.$$ Done!

Answer 2

I have a generalization of the Brahmagupta theorem, but I did not go into the proof process. Definitely valid:

From point $K$ outside a circle $C_1$ we draw the lines $KB$ and $KD$ so that they are perpendicular to each other and intersect the circle $C_1$ at points $A, B, C, D$. The construction is possible only if the distance of $K$ from the center of the circle $C_1$ is less than $R \sqrt {2}$, where $R$ is the radius of the circle $C_1$. Prove that if the line $KM$ is perpendicular to $AD$, then $MB = MC = MK$ will hold.