Given:
In △ABC, m∠CAB = 60° and point D ∈ BC so that AD = 10 in, and the distance from D to AB is 5 in.
The distance from D to AB is 5 in.
Construction:
Take point E on AB and join ED such that ED⊥AB
Proof:
Let E be the point on AB which is 5 in from point D.
DE would be perpendicular to AB because distance is always measure in perpendicular form.
Therefore, ΔAED is a right angle triangle.
where, ∠AED=90° , AD=10 and ED=5
In ΔAED, ∠AED=90°
Using trigonometry identities
[tex]\sin(\angle EAD)=\frac{\text{Perpendicular}}{\text{Base}}=\frac{ED}{AD}[/tex]
[tex]\sin(\angle EAD)=\frac{5}{10}=\frac{1}{2}[/tex]
Therefore, ∠EAD=30°
But ∠CAB=60°
∠BAD=∠CAB-∠EAD=60-30 = 30°
Therefore, ∠BAD=∠EAD=30°
We can say line AD is angle bisector of ∠A. Because it divided ∠A into two equal part.
Hence Proved
