Question

ΔABC is isosceles with AB=AC, P is mid point of BC. if PM is perpendicular to AB and PN is perpendicular to AC, show that PM=PN. View the attached document to get the figure. Also suggest me a simple way to give attachment of a figure

8cd36dc38728816e404378476d294af7.docx

Answer 1

Join AP
AP is the median as P is the midpoint
We know that, median of a triangle divides the triangle into two triangles of equal area.

Therefore,
Area of ΔAPB = Area of ΔAPC
⇒ $\frac{1}{2}*AB*PM = \frac{1}{2}*AC*PN$

But AB = AC (given)

so, PM = PN

Answer 2

You may also prove by showing that triangles BPM and PCN are congruent.

1.Triangle ABC being isosceles, angles b and C are equal.
2.P being the mid-point, sides BP and PC are equal.
3.Angles BPM and CPN are equal, being equal to (90 - Angle B) and (90 - Angle C) respectively.
Hence triangles are congruent and hence PM = PN.