In triangle ABC; ∠A = 60°, ∠C = 40° and bisector of angle ABC meets side AC at point P. Show that BP = CP.

In the given figure, AB ⊥ BE, EF ⊥ BE, AB = EF and BC = DE, then :

1. △ ABD ≅ △ EFC

2. △ ABD ≅ △ FEC

3. △ ABD ≅ △ ECF

4. △ ABD ≅ △ CEF

From the given figure, if ∠A = ∠C, we get :

1. x = 8, y = 16

2. x = -8, y = 16

3. x = 16, y = -8

4. x = 16, y = 8

ABCD is a rectangle. X and Y are points on sides AD and BC respectively such that AX = BY, then :

1. AY ≠ BX

2. △ ABX ≅ △ BYA

3. △ ABX ≅ △ AYB

4. △ ABX ≅ △ BAY