Question

ABCD is a parallelogram. A circle through A, B and C intersects CD produced at E. Prove that AE=AD

Answer 1

the above picture is the prove

Answer 2

AnswEr:

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$\setlength{\unitlength}{1.2 mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(8,40){\line(5,0){35}}\put(38,9){\sf{B}}\put(12,15){\line(5,0){26}}\put(8,40){\line(5,0){35}}\put(12,15){\line(1,5){5}}\put(12,15){\line( - 1,5){5}}\put(38,15){\line(1,5){5}}\put(9,10){\sf{A}}\put(44,42){\sf{C}}\put(18,43){\sf{D}}\put(3,42){\sf{E}}\end{picture}$

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⠀⠀⠀⠀⠀⠀☯ In order to prove that AE = AD i.e. ∆ AED is an Isosceles triangle it is sufficient to prove that $\bf \angle AED = \angle ADE$. Since ABCD is a cyclic quadrilateral.

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$\therefore\;\sf \angle AED + \angle ABC = 180^\circ\qquad\qquad\bigg\lgroup\bf eq\;(1)\bigg\rgroup$

☯ Now, CDE is a straight line

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$\therefore\;\sf \angle ADE + \angle ADC = 180^\circ$

☯ But, $\bf \angle ADC\;and\; \angle ABC$ are opposite angles of a parallelogram.

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$\therefore\;\sf \angle ADC = \angle ABC$

$:\implies\sf \angle ABC + \angle ADE = \angle ADC + \angle ADE$

$:\implies\sf \angle ABC + \angle ADE = 180^\circ\qquad\qquad\bigg\lgroup\bf eq\;(2)\bigg\rgroup$

☯ Now, From equations (i) and (ii), we get

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$:\implies\sf \angle AED + \angle ABC = \angle ABC + \angle ADE$

$:\implies\sf \angle AED = \angle ADE$

☯ Thus, In ∆ AED, we have

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$:\implies\sf \angle AED = \angle ADE$

$:\implies\sf \purple{AD = AE}$