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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


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.

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

But AB = AC (given)

so, PM = PN

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Can you please relate it to tenth standard triangles
Good explanation. ★ ☆
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.
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