The eighth grade mathematics scholars' education edition Volume I answers (1) 14 pages.
1. Solution: ACD=? B.
Reason: Because of the CD? AB,
So △BCD is a right triangle,
? BDC=90? ,
So what? B+? BCD=90? ,
Because again? ACB= 90? ,
So what? ACD+? BCD=? ACB=90? ,
So what? ACD=? B (complementary angles of the same angle are equal).
2. Solution: △ADE is a right triangle,
Reason: Because? C=90 .,
So what? A+? 2=90。 .
Because again? 1= ? 2,
So what? A+? 1=90? .
So △ADE is a right triangle (a triangle with two complementary angles is a right triangle).
Eighth-grade Mathematicians Education Edition Volume 1 Answer (2) Exercise 12.2
Eighth-grade Mathematicians Education Edition Volume I Answer (III) Page 50.
1. Hint: Do? The bisector of AOB intersects with MN at a point, then the point is point P. (Figure omitted)
2. Prove: As shown in figure 12-3-25, the intersection points P are PF and PG respectively, and the pH is perpendicular to the straight lines AC, BC and AB.
The vertical feet are f, g, H g, H.
∵BD is in △ABC? The bisector of ABC's outer corner, point P is on BD. Pg = ph Similarly, PE=PG. PF=PC=PH。