Question

ABCD is a quadrilateral in which bisectors of angle A and angle C meet DC produced at Y and BA produced at X respectively . prove that angle X + angle Y is equal to 1/2 of angle A + angle C .​

Answer 1

Answer:

Proof is given below.

Step-by-step explanation:

Let the quadrilateral be as given in the attached figure.

To prove:

∠ X + ∠ Y = $\frac{1}{2}$ * (∠ A + ∠ C)

From the figure:

AY is the bisector of ∠ DAB => ∠ DAY = ∠ YAB = $\frac{1}{2}$ * ∠ A

CX is the bisector of ∠ BCD => ∠ BCX = ∠ XCD = $\frac{1}{2}$ * ∠ C

Proof:

Consider the quadrilateral XAYC

We know that sum of all angles of a quadrilateral is 360°

=> ∠ X + ∠ XAY + ∠ Y + ∠ YCX = 360°          

=> ∠ X + ∠ XAD + ∠DAY + ∠ Y + ∠ YCB + ∠BCX = 360°      ---(i)

We know that sum of angles on a straight line is 180°

=> ∠ XAD + ∠ DAY + ∠ YAB = 180°

=> ∠ XAD + ∠ A = 180°

=> ∠ XAD =  180° -  ∠ A                                      --(ii)

and ∠ DCX + ∠ XCB + ∠ BCY = 180°

=> ∠ C + ∠ BCY = 180°

=> ∠ BCY = 180° -  ∠ C                                       --(iii)

Substituting the values (ii) and (iii) in equation (i)

=> ∠ X + (180° -  ∠ A) + ∠DAY + ∠ Y + (180° -  ∠ C ) + ∠BCX = 360°

=> ∠ X -  ∠ A + ∠DAY + ∠ Y -  ∠ C  + ∠BCX = 0

=> ∠ X  + ∠DAY + ∠ Y  + ∠BCX = ∠ A + ∠ C

=> ∠ X  + ($\frac{1}{2}$ * ∠ A) + ∠ Y  + ($\frac{1}{2}$ * ∠ C) = ∠ A + ∠ C

=> ∠ X   + ∠ Y    = ∠ A + ∠ C -  ($\frac{1}{2}$ * ∠ A) - ($\frac{1}{2}$ * ∠ C)

=> ∠ X   + ∠ Y    =  $\frac{1}{2}$ * ∠ A + $\frac{1}{2}$ * ∠ C

=> ∠ X   + ∠ Y    =  $\frac{1}{2}$ (∠ A + ∠ C)

Hence proved.

Answer 2

Answer:

In the AABY: ZY + ZA /2 + B = 180° In the ADCX: 4D+2C/2 + X = 180°

Add them to get: (<A + <C)/2 + 2B + <D + 2X+2Y = 360° we know <A + ZB + C + D = 360°

From these two equations, we get:<x +<Y = (<A+ 2C)/2