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Source. Section 4.17, printed pp. 459–460 (published PDF).

Statement. If $T$ is a Hungarian tree in $G$ with inner set $I$, then

$$\nu(G)=|I|+\nu(G-V(T)).$$

Thus a matching on the complement is maximum there exactly when its union with any maximum matching of $T$ is maximum in $G$.

Proof. Combining disjoint maximum matchings of the tree and complement gives at least the stated size, by Section 4.2.

For an arbitrary matching $L$, separate edges lying in the tree, edges lying in its vertex complement, and all remaining edges. Every remaining edge meets an inner vertex: an outer vertex has no neighbor other than an inner vertex of $T$, and edges inside $T$ already go between the two classes. This also handles chords not among the tree edges.

Let $I'$ be the inner vertices met by remaining edges. Their number is at least the number of these edges, as $L$ is a matching. The edges of $L$ that lie in the tree avoid $I'$ and use distinct vertices of $I\setminus I'$, so their number is at most $|I|-|I'|$. There are at most $\nu(G-V(T))$ edges entirely in the complement. Adding the three bounds proves the equality.

A nonmaximum matching in the complement can plainly be improved without changing the disjoint tree matching; the equality proves the converse. $\square$