4合理类比原有图形,灵活改造图形条件原题是由三角形的内心、外心构造出的命题.若将两内心换成垂心,则有图7题6如图7,给定正△ABC,D是边BC上任意一点,△ABD的外心、垂心分别为O1、H1,△ACD的外心、垂心分别为O2、H2.求证:(1)∠H1O1H2=∠H1O2H2;(2)O1H1+O2H2为定值;(3)|S△ABD-S 4 Reasonable analogy of the original graphics, flexible transformation of the graphics condition The original question is a proposition constructed from the inner and outer sides of the triangle. If the two hearts are replaced with the vertical heart, there is Figure 7 Question 6 as shown in Figure 7, given positive △ ABC, D is an arbitrary point on the side BC. The outer and center of circumflex of △ABD are O1 and H1, respectively. The outer and center of circumcenter of △ACD are O2 and H2, respectively. Proof: (1) ∠H1O1H2=∠H1O2H2; (2) O1H1+ O2H2 is a fixed value; (3)|S△ABD-S