Search results
Create the page "Cla" on this wiki! See also the search results found.
- ...</math>, <math>GLN</math> and <math>CEB</math>, <math>GEI</math> and <math>CLA</math>, we get <math>DE \parallel IH</math>, <math>KD \parallel MC</math>,3 KB (502 words) - 22:58, 5 October 2015
- ...h>L</math> is the midpoint of <math>AB</math>, and <math>\angle CLB=\angle CLA=90^\circ</math>. Draw the perpendicular from <math>I</math> to <math>CB</ma16 KB (2,817 words) - 07:39, 5 January 2025
- minor edit by ~CLA13 KB (2,175 words) - 15:16, 12 December 2024
- ...isosceles condition, we have <math>\angle{BKA}=180^{\circ}-2\alpha, \angle{CLA}=120^{\circ}-2\alpha</math> ...express <math>AC^2, AB^2, AC^2+AB^2=2\cdot 14^2(2-\cos \angle{BKA}-\angle {CLA})=2\cdot 14^2(2+\cos(2\alpha)+\cos(60^{\circ}-2\alpha))=38^2</math>18 KB (2,767 words) - 08:57, 18 February 2025