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⊥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

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

127

whether it's diameter, or radius, i believe if i remember correctly from school, every amount has to always add up to 180. so if you take 53, and starting adding, long story short, taking 127 and adding it to 53, gives you 180.