# If AP bisects angle BAC and M is any point on AP.prove that the perpendiculars drawn from M to AB and AC are equal

1
by Krity

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by Krity

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given

AP bisects <BAC & M is any point on AP

proof

<BAM=CAM (given AP bisects)

<AXM= <CYM (90)

AM=AM (COMMON SIDE)

So the triangles AXM is cong. to AYM by AAS

hence,XM=YM (CPCT)

i.e.the perpendiculars are equal