in triangle abc d and f are the midpoints of the AC and ab respectively the altitude of AP to BC intersect EF at Q prove that AQ=QP
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AC=AC(common)
AP=BC(By construction)
EQ=FQ
By sss rule
Therefore,, AQ=QP(By c.p.c.t)
Plss mark at brainlist
AP=BC(By construction)
EQ=FQ
By sss rule
Therefore,, AQ=QP(By c.p.c.t)
Plss mark at brainlist
Answered by
1
mark me as brainliest...
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