Answers

2015-02-23T13:43:18+05:30
GIVEN- A parallelogram ABCD
a circle through A,B,C intesects CD produced at E
TO PROVE- AE=AD
PROOF- angle ADE +
angle ABC =180 ........1 (sum of opp. angle of a cyclic qurd.)
angle ADE +angle ADC=180 ...........2  (LINEAR PAIR)
angle ABC = angle ADC=180 .......3 (opp. angle of a parallogram are equal)
from 1 and 2
angle ADE +angle ADC = angle ABC = angle ADC
angle AED = angle ADE (using 3)
In triangle AED,
angle AED = angle ADE
AE = AD (equal sides have equal angles angles opp. to them)

HENCED  PROVED....



0
please mark it as the best
GIVEN- A parallelogram ABCD
a circle through A,B,C intesects CD produced at E
TO PROVE- AE=AD
PROOF- angle ADE +angle ABC =180 ........1 (sum of opp. angle of a cyclic qurd.)
angle ADE +angle ADC=180 ...........2 (LINEAR PAIR)
angle ABC = angle ADC=180 .......3 (opp. angle of a parallogram are equal)
from 1 and 2
angle ADE +angle ADC = angle ABC = angle ADC
angle AED = angle ADE (using 3)
In triangle AED,
angle AED = angle ADE
AE = AD (equal sides
2015-02-23T14:14:26+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
I think the proof is that    AE = BD,    AD = BE.  There is a mistake in the given question.

See the diagram.
ABCD is a parallelogram.  The circumcircle through A,B, & C intersects extended CD in E.

We know that in a circle, a chord inscribes the same angle at any point on the circumference on the same side.

Let  angle AED = x.    Let angle DAE = y  and  let angle ADB = z.

\angle AED=\angle ABD=x\\As\ AB\ ||and\ DE,\\ AE\ (transversal)\ intersects\ both,\ \angle BAE=\angle AED=x.\\\\Transversal\ DB\ intersects\ parallel\ lines\ AB\ and\ DE,\\.\ \ \ \angle BDE=\angleABD=x.\\\\Chord\ DE\ makes\ \angle DBE\ =y\ and\ \angle DAE=\angle DBE=y\\\\Chord\ AB\ makes\ \angle ADB=z,\ and\ \angle AEB=\angle ADB=z\\Transversal\ BD\ intersects\ parallel\ lines\ AD\ and\ BC.\\Hence,\ \angle ADB=\angle DBC=z\\


Compare the triangles  ADE and BDE.

  DE is the common side.  angle BDE = angle AED.   angle DAE = angle DBE.
  So the triangles are similar and as one side is common,  the triangles are congruent. 

       Hence,  AD = BE   and    AE = BD
             Since,  AD = BC,       AD = BE = BC.

1 5 1