two and (by Prop. 24) the base opposite the greater angle is the greater. But in triangles A B C, DEF the base EF is greater than A C; therefore E F is opposite the greater of the two angles BAC, EDF; that is to say, the angle E D F, to which E F is opposite, is greater than BA C.

XXV.

If from a point without a straight line a number of straight lines be drawn to the former one whereof one is perpendicular, the shortest line is the perpendicular, and of the others the one making a smaller angle with the perpendicular is shorter than one making a larger angle with it; and only two equal straight lines can be drawn from the point to the straight line, viz., at equal angles on either side of the perpendicular.

Let A (Fig. 18) be a point without the straight line DB; AB perpendicular to DB; AC a straight line from A cutting DB in C; AD another similar line, making angle D A B greater than CAB; AE falling on DB on the other side of B, making angle BA E equal to BA C.

Then A B will be the shortest of all the lines A B, AC, AD, etc.; and of the others, AC is shorter than AD, and AE is the only line that can be drawn from A to D B equal to A C.

Because the angle A B C is a right angle, the angle A CB is less than a right angle; wherefore

the side A C is greater than A B, or A B is shorter than any other line drawn from A to D B.

Again, because the angle ACB is less than a right angle, the angle A CD in the triangle A CD is greater than a right angle, and, therefore, the angle ADC is less than a right angle, or the angle A CD of the triangle A C D is greater than ADC; wherefore the side AD is greater than the side A. C.

And because the angle BAE is equal to the angle B A C, and the angle A BE is equal to the angle ABC (being both right angles), the triangles A B C, ABE have two angles of the one equal to two angles of the other, and the side A B common to the two. Therefore the triangles ABC, A B E are equal, and the side A E is equal to A C, and every other line on the other side of B from C is (by the present Prop.) either greater than A E or less; wherefore A E is the only line that can be drawn from A to D B equal to A C.

Cor. 1.-If a straight line be shorter than any other that can be drawn from a point without a straight line to that line, the former straight line is perpendicular to the latter.

Cor. 2.-The perpendicular on the base of an isosceles triangle bisects the angle at the vertex.

XXVI.

The straight line joining any two points in the

circumference of a circle falls wholly within the circle.

Let AB (Fig. 19) be a straight line joining A and B points in the circumference of a circle whose centre is C. AB will fall wholly within the circle. Let CA, CB be radii of the circle at the points A and B; CD perpendicular to A B. Then (Prop. 25, Cor. 2) CD bisects the angle B A C, and every straight line from C to the straight line AB, between CA and C B, will be less than CA or CB (Prop. 25), and therefore every point in AB falls within the circle.

XXVII.

If a straight line joining any two points in the circumference of a circle be bisected, and a perpendicular erected at the point of bisection, the centre of the circle lies in the perpendicular.

Let A and B (Fig. 20) be points in a circle, and let the straight line AB be bisected in D; DC perpendicular to AB. The centre of the circle will be in the line DC.

Let F be any point within the circle not in the straight line DC. Join FA, FB, of which FA cuts DC in E. Join E B. Then E A is equal to EB (Prop. 26), and AF is equal to A E and E F, that is, to E B and E F. But EB and EF are together greater than FB, the third side of the triangle E B F; therefore A F is greater than F B.

Wherefore F is not the centre of the circle, or, in other words, the centre of the circle does not lie in any point without the perpendicular C D, that is, it lies upon it.

XXVIII.

If a straight line cut a circle, it shall cut it again at a point at an equal distance on the other side of the perpendicular from the centre and at no other point.

Let the straight line A B (Fig. 19) cut the circle whose centre is C, at the point A, and let CD be perpendicular upon A B. A B shall cut the circle again at a point B so that DB is equal to DA, and at no other point.

Join CB, CA. Then because DA is equal to DB, and C D is common to the triangles CDA, CD B, and the angles CD A, C D B are both right angles, the sides DA, CD are respectively equal to the two DB, DC, and the included angle CDA is equal to the angle CDB; wherefore the triangles are equal and the base CB is equal to CA, or B is a point in the circle, and the line AB (Prop. 27) falls wholly within the circle.

And because no other line besides CB can be drawn from C to AB equal to CA (Prop. 26), no other point in A B coincides with any point in the circle; or, in other words, the straight line AB cuts the circle in no other point.

XXIX.

The perpendicular at the extremity of a diameter falls without the circle, and no straight line can be drawn between that and the circle so as not to cut the circle.

Let A B (Fig. 21) be the diameter of a circle; C, the centre; BE perpendicular to BC. BE will fall wholly without the circle, and no straight line can be drawn between BE and the circle so as not to cut the circle.

Let CF be a line joining C and any point F in BE. Then, because CBF is a right angle, the angle C F B is less than a right angle; wherefore CF is greater than CB, or the point F lies without the circle.

Next let B D be any other straight line through B, and let CD be perpendicular to BD. Then because CDB is a right angle, CDB is greater than CBD, and CB is greater than CD; or the point D falls within the circle. And any line from C to DB, beyond CB, will be greater than C B ; wherefore every part of the line DB on the other side of B will fall without the circle. Therefore the circle intersects D B in B.

Cor.-If a straight line touch a circle, the centre of the circle lies in the perpendicular to the touching line at the point of contact.