Vì `IK //// AC=>AE //// EB; AE //// BK`

Xét `2 triangle KHB, AHE` có `AE //// EB`

`=>(AH)/(KH)=(EH)/(HB)`

Xét `2 triangle IHB, AHE` có `AE //// EB`

`=>(AH)/(IH)=(EH)/(HB)`

`=>(AH)/(KH)=(EH)/(HB)=(AH)/(IH)`

`=>(AH)/(KH)=(AH)/(IH)`

`=>HI=HK` (đcpm)

`- \text(Lamtoanbangcatinhmang) -`