∆ ABC is an equilateral triangle. seg AD perpendicular to side BC which of the following relations is true ? *
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Answer:
△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
Step-by-step explanation:
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