How protective diode protects transistor from breakdown?

Please, explain the process of breakdown; how exactly this protective diode protects the transistor?

In a Horowitz & Hill book, "The Art of Electronics", 2nd edition, in "Chapter 2 - Transistors" (page 68), i read the following:

2. Always remember that the base-emitter reverse breakdown voltage for silicon transistors is small, quite often as little as 6 volts. Input swings large enough to take the transistor out of conduction can easily result in breakdown (with consequent degradation of hFE unless a protective diode is added (Fig. 2.10).

Can't figure out how this diode protects the transistor from breakdown if the current goes only in one direction in this diode.

^{simulate this circuit – Schematic created using CircuitLab}

Answer

The diode is in place to protect the transistor from reverse \$V_{BE}\$ breakdown. If you reverse-bias the base-emitter junction by taking the transistor's emitter to ~\$6\rm{V}\$ more positive than the base, it will breakdown and begin conducting.

This reverse breakdown damages the base-emitter junction, causing a degradation in \$h_{FE}\$.

By placing a diode as shown, the reverse-bias voltage is limited to ~\$0.7\rm{V}\$; attempts to apply more voltage will be futile, since the diode will conduct a lot of current and prevent increased voltage. This protects the transistor's base-emitter junction.