Let DEF be a circle, inscribed in the right-angled triangle ABC. The difference between AC, and AB, BC, is equal to the diameter of the circle.

Find O the centre, and join OD, OE. Then the angles at D, B, and E being

D

B I

right angles, and OD=OE, OB is a square; and DB, BE are equal to OD, OE, i. e. are together equal to the diameter of the circle. Now (Eucl. iii. 36. Cor.) CE= CF, and AD=AF; i. e. AC is equal to AD and CE; whence it is less than the sides containing the right angle, by DB and BE, or by the diameter of the circle.

(4.) If a semicircle be inscribed in a right-angled triangle, so as to touch the hypothenuse and perpendicular, and from the extremity of its diameter a line be drawn through the point of contact to meet the perpendicular produced; the part produced will be equal to the perpendicular.

Let the semicircle ADE touch the hypothenuse BC of the right-angled triangle ABC in D, and the perpendicular in A; and from E let ED be drawn to meet AB produced in F; AB=BF.

B

Join AD. Since ADE is a right angle, ADF is also a right angle, and .. equal to DAF, DFA together. But DAF is equal to BDA, since BD =BA, being tangents from the same point B without the circle; and .. the angle BFD = BDF, and BF= BD=BA.

(5.) If the base of any triangle be bisected by the diameter of its circumscribing circle, and from the extremity of that diameter a perpendicular be let fall upon the longer side; it will divide that side into segments, one of which will be equal to half the sum, and the other to half the difference of the sides.

Let the base BC of the triangle ABC be bisected in E, by the diameter of the circumscribing circle ACD; and from D draw DF perpendicular to AB the longer side; BF will be equal to half the sum, and AF to half the difference of AB, BC.

Join DA, DB, DC; and make BG = BC; join DG. Since BG=BC, and BD is common, and the angle GBD=CBD, since the arc AD=DC; :. DG= DC=DA, and DF is at right angles to AG, .. AF= FG. Whence the sum of AB and BC is equal to AG and 2BG, i. e. to 2 BF; and the difference of AB and BC is equal to the difference of AB and BG, i. e. to 2 AF.

(6.) The same supposition being made, as in the last proposition; if from the point, where the perpendicular meets the longer side, another perpendicular be let fall on the line bisecting the vertical angle; it will pass through the middle of the base.

The same construction being made as before; (see last Fig.) let FH be drawn perpendicular to BD, which bisects the vertical angle; FH will pass through E.

Because CB = BG, and the angle CBD = GBD,

.. BD is perpendicular to CG; and .. FH is parallel to CG. But since AF= FG, .:. (Eucl. vi. 2.) AC is bisected by FH; which.. passes through E.

(7.) If a point be taken without a circle, and from it tangents be drawn to the circle, and another point be taken in the circumference between the two tangents, and a tangent be drawn to it; the sum of the sides of the triangle thus formed is equal to the sum of the two tangents.

From a given point D let two tangents DA, DB be drawn ; and to C any point in the circumference between them, let a tangent ECF be drawn. The sum of the sides of the triangle is equal to the two tangents DA and DB.

E

B

Since AE= EC, and FC=FB, .. DE, EF, FD together are equal to AD and DB together. In the same manner, if through any other point in the arc ACB a tangent be drawn, it will be equal to the two segments of DA, DB intercepted between it, and the points of contact A and B; and the three sides of the triangle so formed will be equal to DA, and DB together.

(8.) Of all triangles on the same base and between the same parallels, the isosceles has the greatest vertical angle.

Let ABC be an isosceles triangle on the base AC, and between the parallels AC, BD. It has a greater vertical angle

B

than any other triangle ADC on the same base, and between the same parallels.

About ABC describe a segment of a circle ABC; and since B is the middle point of the arc, and BD is parallel to AC, BD is a tangent at B. Let the arc cut AD in E; join EC. Then the angle ABC = AEC, and .. is greater than ADC.

COR. Of all triangles on the same base and having the same vertical angle, the isosceles is the greatest. For the triangle AEC has the same vertical angle with ABC, and ABC= ADC on the same base and between the same parallels; but ADC is greater than AEC, .. ABC is greater than AEC.

(9.) If through the vertex of an equilateral triangle a perpendicular be drawn to the side, meeting a perpendicular to the base drawn from its extremity; the line intercepted between the vertex and the latter perpendicular is equal to the radius of the circumscribing circle.

Let BE perpendicular to AB meet AE, which is perpendicular to the base AC, in E; BE is equal to the radius of the circle described about ABC.

I G

F

Draw BF, CG perpendicular to the sides; and produce CG to H. Then CI is equal to the radius of the circle described about ABC; and EBIH is a parallelogram. And since CF is equal to FA, (Eucl. vi. 2.) CI is equal to IH, i. e. to the opposite side BE; and .. BE is equal to the radius of the circumscribing circle.

(10.) If a triangle be inscribed in a semicircle, and a perpendicular drawn from any point in the diameter, meeting one side, the circumference, and the other side produced; the segments cut off will be in continued proportion.

Let ABC be a triangle in the semicircle ABC; and from any point D in the diameter, let DF be drawn perpendicular to AD, meeting BC, the circumference, and AB produced, in E, G, F; DE DG:: DG: DF.

For the angles at E being equal, and the angles at Bright angles, .. the angle ECD is equal to BFD; and the angles at D are right angles; .. the triangles EDC, ADF are similar, and therefore

DF

but DA

DA :: DC: DE,

DG:: DG: DC;

.. ex æquo DF: DG :: DG: DE.

(11.) If a triangle be inscribed in a semicircle, and one side be equal to the semi-diameter; the other side will be a mean proportional between that side and a line equal to that side and the diameter together.

Let ABC be a triangle inscribed in the semicircle, and let BC be equal to the semi-diameter; then will

A

BC BA: BA: BC+CA.

Produce AC to D, making CD equal to the semidiameter. Take O the centre. Join BD, BO. Since