Dictionary of junior high school mathematics professional terms

Solution: We need to use twice triangular congruence.

Proof: from the meaning of the question, BD is the bisector of ∠EBF, so ∠EBD=∠FBD. & lt 1 & gt；

Known: |BE|=|BF|. & lt2 & gt

Through < 1 >& lt2> and BD on male * * * side, it is obtained that △EBD is equal to △FBD. So ∠ MDP = ∠ NDP <; 3 & gt

Then you know that PM is perpendicular to DE and PN is perpendicular to DF. So Rt∠PMD=Rt∠PND. & lt4 & gt

Again, from < 3 >& lt4> and public PD, it is obtained that △PMD is equal to △PND. So |PM|=|PN|.

Prove completion. Welcome questions!