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PROPOSITION X. The straight line joining the centres of two intersecting circles bisects their common chord at right angles.

Let FG be the common chord of the intersecting circles PFG, QFG.

The straight line joining their centres shall bisect FG at right s.

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Then :: the chord FG of PFG is bisected at right 2 s by PQ;

.. centre of O PFG is in PQ; (111. 2) so also in like manner is the centre of O QFG; .. the straight line joining the centres of the Os PFG, QFG bisects FG at right 2 s.

PROPOSITION XI.

If one circle touch another internally the straight line joining their centres shall pass through the point of contact.

Let the o CQ, whose centre is B, touch the o CR, whose centre is A, internally at C.

Then shall AB produced pass through C.

For, if not, if possible, let it fall otherwise, as ABQR.

Then AB, BC are together > AC, (1. 16)

but BC = BQ and AC = AR.
.. AB, BQ are together > AR, which is impossible.

AB produced must pass through C.

PROPOSITION XII. If two circles touch externally the straight line joining their centres shall pass through the point of contact.

Let H, K be the centres of two circles AB, AC touching externally in A. · Then shall the straight line joining HK pass through A.

For, if not, if possible, let it fall otherwise, as HBCK.

Join HA, KA.

Then HA, AK are together > HK, (1. 16)

but HA = HB and AK = CK; .: HB, CK are together > HK, which is impossible. .. If two circles, &c.

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DEFINITIONS. A segment of a circle is the figure contained by a straight line and the part of the circumference which it cuts off.

An angle is said to stand upon the circumference intercepted between its arms.

PROPOSITION XIII. The angle at the centre of a circle is double of the angle at the circumference upon the same arc.

Let AB be an arc of a o of which C is the centre,

ACB an angle upon AB at the centre,

APB an angle upon AB at the Oce.
Then shall - ACB be double of L APB.

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2ndly. Let C lie within the LAPB.

Join PC, and produce it to meet the Oce in X.

Then ACX is double of L APX,

and LBCX is double of . BPX; .. the whole LACB is double of LAPB.

3rdly. Let C lie without the LAPB.

Join PC, and produce it to meet the Oce in Y.

Then - BCY is double of BPY,

and LACY is double of LAPY;
.: remaining - ACB is double of 1 APB.

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