Question

Can anyone help me out with question no.16?

Answer 1

ang. BMN + ang. MND = 180° ( Co Interior Angles)
and,
ang. BMN = ang. MND (co interior angles)
so, let ang. BMN = ang. MND = x.
x + x = 180°
2x = 180°
x = 180/2
x = 90°
ATQ,
MP bisects ang. BMN
so, ang. BMP = ang. PMN = 1/2 of ang. BMN
= 1/2 of 90 = 45°
so, ang. BMP = ang. PMN = 45°
similarly,
NP bisects ang. MND
so, ang. MNP = ang. PND = 45°
In triangle MPN,
ang. PMN + ang. MNP + ang. MPN = 180° (A.S.P. of traingle)
45°+45°+ ang. MPN =180°
ang. MPN = 180°-90°
ang. MPN = 90°
Hence Proved.