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Apm is 0585xhdey we ygg with afhir dihc ayhj
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- construction: draw a line from Mid of A to mid of CB and MP (where they intersect each other) lets take that point ad O
Step-by-step explanation:
consider TRIANGLE AMO and BOM
angle M = angle B (90 degree)
side AO = side AO (common)
by this triangles are equal
SO, AM =AB
AM= half AC
AB= half AP
so half AC = half AP
half cancled ( common in both )
so AC= AP
(hence it is proved that AB×AP=AC×AM)
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