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?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial