y*w
发帖数: 238 | |
l*****a
发帖数: 375 | |
N***l
发帖数: 52 | 3 我算出来20度。
自D引BC垂线交BC于F交AE于G.
延长CG交BD于H
则CGH三点均在AB中垂线上。
再证明三角形DHA相似于DGE,用几次角分线定理和
三角形GEF是一内角为30度的直角三角形。
做不出来，有答案吗？
【在 l*****a 的大作中提到】
: 做不出来，有答案吗？
|
e**********n
发帖数: 359 | |
e**********n
发帖数: 359 | 5 Problem 1.
20 degrees, solution:
1. Find F on EB so that DF // AB.
2. Extend AF to G so that AC = AG, now triangle ABC and triangle CGA are
identical.
3. Extend GE to H on DC. It’s easy to see that the angle HEA = 30 degrees.
4. Next is to show angle HED = 10 degrees, which is done by noticing the
triangle HEF is equilateral, and the triangle DEF is isosceles.
5. x = angle HEA - angle HED = 20 degrees.
Problem 2.
30 degrees, solution:
1. Find F on DC so that FE // AB, FB intersect AE at G, a |
t**i
发帖数: 688 | 6 Could you please highlight "则CGH三点均在AB中垂线上"? Thanks. |
N***l
发帖数: 52 | 7 容易得出角DGA=60，所以DGBA四点共圆
故DBG=10，从而GB平分DBE。而DG平分CDB，所以G必为
三角形BCD内心，则，CG平分ACB.
【在 t**i 的大作中提到】
: Could you please highlight "则CGH三点均在AB中垂线上"? Thanks.
|
t**i
发帖数: 688 | 8 The new bug of BBS is really making troubles to view the last post. |
t**i
发帖数: 688 | 9 Thanks, Nobel and Eventhorizon. Both are good problem solver. |
