In △ABC, m∠CAB = 60° and point D ∈ BC so that AD = 10 in, and the distance from D to AB is 5in. Prove that AD is the angle bisector of ∠A.

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⊥ABProof: 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=5In 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