HMMT 十一月 2021 · 冲刺赛 · 第 6 题
HMMT November 2021 — Guts Round — Problem 6
题目详情
- [6] Let ABCD be a parallelogram with AB = 480, AD = 200, and BD = 625. The angle bisector of ∠ BAD meets side CD at point E . Find CE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2021, November 13, 2021 — GUTS ROUND Organization Team Team ID#
解析
- [6] Let ABCD be a parallelogram with AB = 480, AD = 200, and BD = 625. The angle bisector of ∠ BAD meets side CD at point E . Find CE . Proposed by: Guanpeng Xu, Joseph Heerens Answer: 280 Solution: D E C 200 625 A B 480 ◦ ∠ BAD First, it is known that ∠ BAD + ∠ CDA = 180 . Further, ∠ DAE = . Thus, as the angles in tri- 2 ∠ BAD ◦ angle ADE sum to 180 , this means ∠ DEA = = ∠ DAE. Therefore, DAE is isosceles, making 2 DE = 200 and CE = 280 .