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In ΔABC, the perpendicular from vertices A and B on their opposite sides meet (when produced) the circumcircle of the triangle at points D and E respectively.

Prove that arc CD = arc CE

Somebody, please help me with this proof.

Note: You are not allowed to use trigonometry

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,651

Hi Agnishom

For arc CD = arc CE you want angle DAC = angle CBE

See diagram. Where the altitudes meet the sides I've named those points F and G

So what can you say about triangles AFC and BGC ?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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They have one common angle and one right angle so the other angle must be the same.

thanks but I need a way to say that if angle angle DAC = angle CBE, then CD = CE

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,651

Let O be the point at the centre of the circle.

angle DOC = 2 x angle DAC (angle properties of a circle)

similarly angle COE = 2 x angle CBE

OD = OC = OE = radius

So triangles DOC and COE are congruent => straight line DC = straight line CE => arc DC = arc CE

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Okay Got it Thanks!

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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