Question 209023
<pre><font size = 4 color="indigo"><b>
Given: Triangle ABC is isosceles; 
       Line segment CD is the altitude to the base AB

To prove: Angle ACD = angle BCD

{{{drawing(267,267,-2,2,-1,3, triangle(-1,0,0,2.5,0,0),
triangle(1,0,0,2.5,0,0), locate(-1,0,A),locate(0,0,D),
locate(1,0,B),locate(0,2.7,C)


   )}}}


angle A = angle B       Base angles of isosceles triangle

angle ADC = 90°         An altitude of a triangle is 
                        perpendicular to a side

angle BDC = 90°         An altitude of a triangle is 
                        perpendicular to a side

angle ADC = angle BDC   Both are right angles 

Angle ACD = angle BCD   If two angles of one triangle are
                        equal to two angles of another
                        triangle, the third angles are 
                        equal.

 Edwin</pre>