# How do you prove?

## If a circle with centre O has two chords DE and FG which intersect at point H then prove that $\angle D O F + \angle E O G = 2 \angle E H G$?

Dec 16, 2017

Given

A circle with centre O has two chords DE and FG which intersect at point H .

RTP
we are to prove that $\angle D O F + \angle E O G = 2 \angle E H G$

Construction : D and G are joined.

Proof

Now the central $\angle E O G$ and peripheral $\angle E D G$ are on the same arc GE,

So $\angle E O G = 2 \angle E D G \ldots . . \left[1\right]$

Similarly the central $\angle D O F$ and peripheral $\angle F G D$ are on the same arc FD,

So $\angle D O F = 2 \angle F G D \ldots . . \left[2\right]$

Adding [1] and [2] we get

$\angle D O F + \angle E O G = 2 \angle F G D + 2 \angle E D G$

$= 2 \left(\angle F G D + \angle E D G\right)$

$= 2 \left(\angle F G D + \angle H D G\right)$

$= 2 \angle E H G$
(exterior angle of a triangle is equal to sum of two of the remote interior angles of the triangle.)