Question

正方形ABCD中,EH⊥HG于点P,是说明EF=HG.

Answer 1

过A作AJ∥EF交BC于J,过B作BK∥HG交CD于K,令AJ与BK的交点为M.

∵ABCD是正方形,∴AE∥JF,又AJ∥EF,∴AEFJ是平行四边形,∴AJ＝EF.

∵ABCD是正方形,∴BH∥KG,又BK∥HG,∴BHGK是平行四边形,∴BK＝HG.

∵AJ∥EF、BK∥HG、EF⊥HG,∴AJ⊥BK.

∵ABCD是正方形,∴AB⊥BJ,结合证得的AJ⊥BK,得：∠BAJ＝∠CBK.［同是∠AJB的余角］

∵ABCD是正方形,∴AB＝BC、∠ABJ＝∠BCK=90°,又∠BAJ＝∠CBK,∴△ABJ≌△CBK,

∴AJ＝BK,结合证得的AJ＝EF、BK＝HG,得：EF＝HG.