angle AXD + angle CAX = angle ACD [External property of a triangle]
angle AXD + (1/2) arc BC = (1/2) arc AD [The relation between arc and incribed angle]
angle AXD = (1/2) (arc AD - arc BC) [Taking arc BC to the R.H.S.]
Prove that: angle AXD = 1/2 (arc AD - arc Bc)
angle AXD + angle CAX = angle ACD [External property of a triangle]
angle AXD + (1/2) arc BC = (1/2) arc AD [The relation between arc and incribed angle]
angle AXD = (1/2) (arc AD - arc BC) [Taking arc BC to the R.H.S.]
Prove that: angle AXD = 1/2 (arc AD - arc Bc)
S.E.E. Solution
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Soln: Join AC then angle AXD + angle CAX = angle ACD [External property of a triangle] angle AXD + (1/2) arc BC = (1/2) arc AD [The ...
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