Cho A=(2;0),B(0;2),C=(0;7). Tìm tọa độ đỉnh thứ tư D của hình thang cân ABCD.
Để tứ giác ABCD là hình thang cân, ta cần có một cặp cạnh đối song song không bằng nhau và cặp cạnh còn lại có độ dài bằng nhau.

Trường hợp 1:
                                   
ABCDABCDAD=BC
ABCDABCDCD=kAB,k1
(x0,y7)=(2k;2k){x=2ky=2k+7
Từ AD=(x2;y)
AD=(x2)2+y2BC=(0;5)BC=5
Vậy: AD=BC(x2)2+y2=25
Thế vào: (2k2)2+(2k+7)2=25
4k2+8k+4+4k2+28k+49=25
8k2+36k+28=02k2+9k+7=0
k=1 hay k=72. Chọn k=72 thì D(7;0)

Trường hợp 2: ADBC,ADBCAB=CD
Làm tương tự ta được D=(2,9).

Thẻ

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