# ABCD is a cyclic quadrilateral whose diagonal intersect at T.IF angled(BAD) + angle(BTC) = 180 degree, prove that angle BAC = 1/2{angle(ADC)}.Thankyou if you answer it. Please, can anybody answer...

ABCD is a cyclic quadrilateral whose diagonal intersect at T.IF angled(BAD) + angle(BTC) = 180 degree, prove that angle BAC = 1/2{angle(ADC)}.

Thankyou if you answer it.

Please, can anybody answer it in any way?

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A cyclic quadrilateral has several properties. Among them are:

(1) Opposite angles are supplements

(2) The angle between a side and a diagonal is congruent to the angle formed by the opposite side and the other diagonal.

** In quad ABCD where the diagonals meet at T, `m/_BAC = m/_BDC` ; `/_BAC` formed by `bar(AB),bar(AC)` and `/_BDC` formed by the side opposite which is `bar(DC)` and the other diagonal `bar(DB)`

The proof:

We are given that `m/_BAD+m/_BTC=180` Then

`m/_BAC+m/_CAD+m/_BDC+m/_ACD=180`

** The first two from the angle addition postulate; the second two because the exterior angle of a triangle is equal in measure to the sum of the remote interior angles.

Since the quad. is cyclic we have `m/_BAD+m/_BCD=180` or

`m/_BAC+m/_CAD+m/_BCA+m/_ACD=180`

From these we get `m/_BDC=m/_BCA`

From property (2) we have `m/_BCA=m/_BDA` .

Using substitution we get `m/_BDC=m/_BDA`

From property (2) we have `m/_BAC=m/_BDC`

So `m/_BAC=1/2(2m/_BDC)=1/2(m/_BDC+m/_BDA)`

And `m/_BAC=1/2m/_ADC` as required.