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In a triangle ABC, AC=BC and angle ACB=40. The vertices A,B and C lie on the...

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In a triangle ABC, AC=BC and angle ACB=40. The vertices A,B and C lie on the circumference of a circle. D is a point outside this circle such that DC is perpendicular to BC, BC=DC and the points B and D are on the same side of the line AC. DA cuts the circumference of the circle at E. Find the size of angle DCE.

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We are given `bar(AC) cong bar(BC)` with `Delta ABC` inscribed in a circle. `bar(DC)` is drawn on the same side of `bar(AC)` as B such that `bar(DC) _|_ bar(BC)`and `bar(DC) cong bar(BC)` . Then `bar(DA)` intersects the circle at E; we are to find the measure of `/_DCE` :

`Delta ABC` is isosceles with vertex angle `40^@` so the base angles `/_A cong /_B = 70^@` .

`Delta ACD` is isosceles with vertex angle `130^@` (`/_DCB=90^@` plus `40^@` from `/_ACB` ) so `m/_CAD=m/_CDA=25^@` .

Then `m/_DAB=45^@` (70-25=45)

The measure of arc EBA is 170. (The measure of an inscribed angle is half the measure of its inscribed arc.) The measure of arc EB=90 and the measure of arc BA is 80.

Then `m/_ECB=45^@` (half of arc EB)

`m/_ACD=130^@;m/_ACB=40^@;m/_BCE=45^@` which implies that `m/_DCE=45^@`

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`m/_DCE=45^@`

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