This space is for community interaction. Solutions posted here are provided by our visitors.
Ang MBE = ang EAM = α ( have same arc EFM )ang EAH = ang EDH ( have same arc EH, AEHD concyclic )α = ß
[Here I am using a for alpha and b for beta]From Problem 73, we know that AEHD is a cyclic quadSo <EAH=<EDH=b (< in same seg.)Consider triangle AEBb+<AEB+<MAB+<EBA=180b=180-<AEB-<MAB-<EBA-------(1)Consider triangle MAB<MAB+<AMB+<EBA+a=180a=180-<AMB-<MAB-<EBA---------(2)Since <AEB=<ABM (< in same seg)So (1)=(2)
Share your solution or comment below! Your input is valuable and may be shared with the community.
Ang MBE = ang EAM = α ( have same arc EFM )
ReplyDeleteang EAH = ang EDH ( have same arc EH, AEHD concyclic )
α = ß
[Here I am using a for alpha and b for beta]
ReplyDeleteFrom Problem 73, we know that AEHD is a cyclic quad
So <EAH=<EDH=b (< in same seg.)
Consider triangle AEB
b+<AEB+<MAB+<EBA=180
b=180-<AEB-<MAB-<EBA-------(1)
Consider triangle MAB
<MAB+<AMB+<EBA+a=180
a=180-<AMB-<MAB-<EBA---------(2)
Since <AEB=<ABM (< in same seg)
So (1)=(2)