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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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'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: 7,979

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'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: 7,979

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

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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