Question

In triangle xyz, xy >xz and p is any point on the side yz. Prove that xy >xp

Answer 1

In a triangle XYZ,  XY > XZ where P is any point on the side YZ.

• We can prove it by assuming a circle C1 of radius XY and centre as X

                and another circle C2 with radius XZ and centre as X.

• From the figure it is sure that C1 is bigger than C2,

                   Therefore, ⇵ XY will be greater than  ⇵ XZ.

• Another way of proving is Assume 2 triangles, XYP and XPZ,

                              ∠ XPZ  ≥  ∠ XYP

Therefore, line opposite to that will behave in the same manner,

                     i.e,            XY > XZ

Answer 2

xy > xp , In triangle xyz xy greater than xz p is any point on side yz

Step-by-step explanation:

xy > xz

as angle opposite to larger side is greater

=> ∠z > ∠y

∠ypx = ∠p

∠ypx  = ∠pxz + ∠z

=> ∠p > ∠z  

∠z > ∠y

=> ∠p  >  ∠z  > ∠y

=> ∠p  >   ∠y

in Δxyp

∠p  >   ∠y

=> xy > xp

QED

proved

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