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lukranit [14]
3 years ago
8

Let triangle ABC have bisectors BD, CE intersect at O. Through point A draw perpendicular BD to CE, intersect BC in order at N a

nd M. is called the foot of the perpendicular from O to BC, prove that
a. M is symmetric to A over CE, N is symmetric vs A over BD
b. M is equal to N through OH
Mathematics
1 answer:
mote1985 [20]3 years ago
6 0

Answer:

12 i tink

Step-by-step explanation:

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