Question

如圖1-4-17,在正方形ABCD的邊BC和CD上取點H和M,且==,AH和BM相交于點P,求證:AP =9PH.

圖1-4-17

Answer 1

思路分析:由= =,容易證明△ABH∽△BCM,從而不難推出BP⊥AH,在△ABH中,由=,考慮射影定理確定答案.

證明:在正方形ABCD中,∵= =,?

∴= =.?

∴= =.?

又∠ABH =∠C =90°,?

∴△ABH∽△BCM,∠PBH =∠BAH.?

又∵∠BAH+∠BHA =90°,?

∴∠PBH+∠BHP =90°,即BP⊥AH.?

在Rt△ABH中,設BH =k,則AB =3k, .?

∴AB²=AP·AH,BH²=PH·AH.?

∴= = =.?

∴AP =9PH.