Question

In equilateral triangle abc if ad is a median then prove that angle adc=90

Answer 1

in equilateral triangle ABC , if AD is the median then ,
  BD = CD
  so, consider triangles ABD and ACD,
           AB = AC (sides of equilateral triangle)
          angleABD = angleACD (60 equal angles of equilateral triangle)
          BD = CD (from above)
       so, by SAS congruence criteria , triangle ABD =~ triangle ACD
       so,      angleADB = angleADC ...........................(1)
       but angleADB + angleADC = 180  (supplementary angles)..............(2)
         so, from eqs (1) and (2) , we get,
           ADB + ADB = 180
   so,  2angleADB = 180
so angle ADB = 180/2 => 90
therfore angleADB = angleADC = 90
                                     hence proved

Answer 2

Answer:

here is answer dear

Step-by-step explanation:

ABC is a equilateral triangle.

Thus, AB=BC=AC and∠ABC=∠BAC=∠ACB=60

∠ADB=∠ADC=90

In △ABD,∠BAD+∠ABD+∠ADB=180

⇒∠BAD=180−90−60=30

Similarly for △ACD,∠CAD+∠ACD+∠ADC=180

⇒∠CAD=180−90−60=30

Now, In △ABD and △ACD

AB=AC

∠BAD=∠CAD

AD=AD(Common side)

Thus, △ABD and △ACD are congruent.(SAS)

Therefore BD=DC=

2

1

BC=

2

1

AB

Now, △ABD is a right angled triangle.

Therefore

AB

2

=BD

2

+AD

2

⇒AB

2

=(

2

1

AB)

2

+AD

2

⇒AB

2

=

4

AB

2

+AD

2

⇒AB

2

−

4

AB

2

=AD

2

⇒

4

4AB

2

−AB

2

=AD

2

⇒3AB

2

=4AD

2