Question

if ABCD is a quadrilateral such that AB||CD and AD=BC,prove that angle A=angle B..
SOLVE WITHOUT EXTENDING THE FIGURE ..
DO NOT EXTEND THE FIGURE ..
DO THIS WITHOUT EXTENDING THE FIGURE U WILL BE SURELY GET FOLLOWED ,MARKED BRAINLIEST AND MANY MORE REWARDS​

Answer 1

31

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find the point of X axis which equidistant from the point(3,4) (1-3)

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Let the point be p (x,0) on x axis which is equidistant from the points (3,4) &(1,-3)

on x-axis value of y is zero and on y axis value of x is zero.

(3,4)•A──────────P(x,0)─────────B•(1,-3)

$\bold{ {(PA)}^{2} = {(PB)}^{2}}$

By using distance formula we can solve and find the point' P'

$\bold{\boxed{D = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }}}$

${(PA)}^{2} = \sqrt{ {(x - 3)}^{2} + {(2 - 4)}^{2} }$

$= > \sqrt{ {x}^{2} + 9 - 6x + 16}$

${(PB)}^{2} = \sqrt{ {(1 - x)}^{2} + {( - 3 - 0)}^{2} }$

$= > \sqrt{1 + {x}^{2} - 2x + 9 }$

${(PA)}^{2} = {(PB)}^{2}$

Squaring both sides to remove root :

${ (\sqrt{{x}^{2} + 9 - 6x + 16 } })^{2} = {( \sqrt{1 + {x}^{2} - 2x + 9 }) }^{2}$

$= > {\cancel{{x}^{2}}} + 9 - 6x + 16 = 1 +{\cancel{ {x}^{2}}} - 2x + 9$

$= >9 - 6x + 16 = 1 - 2x + 9$

$= > {\cancel{9 }}- 6x + 16 - 1 + 2x{\cancel{ - 9}}$

$= > - 4x + 15 = 0$

$= > - 4x = - 15$

$\bold{\red{= > x= \frac{15}{4}}}$

Therefore,the point which is equidistant from x axis is (15/4,0)

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