maths

Asked by lakshitas | 27th Feb, 2010, 05:31: PM

Expert Answer:

ATQ,

angle CAP and CBP become angles in the same segment, so they are equal.

so

angle CAP=angleCBP

but angle CBP=B/2... given

so

angle CAP=B/2

Join C ,P.

So ABCP becomes a cyclic quadrilateral,

angle QPC= angle ABC.. as ext angle of a cyclic quadri. is equal to int opp angle.

angle ACP=angle ABP.. angles in the same seg are equal.

 but angle ABP=B/2

so

angle ACP=B/2

angle ACQ=180-angle ACB=180-angle ABC=180-B... since AB=AC given,

i.e.

angle ACP+angle PCQ=180-B

[B/2]+angle PCQ=180-B

so

PCQ=180-[3B/2]

Now in triangle PCQ,

angle QPC+angle PCQ+angle Q=180

so substituting the values, we get,

180-[3B/2]+B+angle Q=180

So,

 angle Q=B/2,

thus in triangle ACQ,

Angle CAQ=B/2

angle Q=B/2

so the triangle is isoceles

 and  so

AC=CQ

hence proved

Answered by  | 28th Feb, 2010, 06:02: PM

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