Let ABC be an isosceles triangle, having the side AB equal to AC. From B and A let fall perpendiculars BD, AE; then will BF: AC :: BE : EA.

B

E

C

Since the angles BDA, AEC are right angles, and the angle DAF common to the two triangles FAD, EAC, .. the triangles are similar. But the triangle BFE is similar to AFD, and .. to EAC; whence BF BE AC: AE,

:

and BF: AC :: BE: EA.

(10.) If from any point in the base of an isosceles triangle lines be drawn to the opposite sides, making equal angles with the base; the triangles formed by these lines, the segments of the base, and the lines joining the intersections of the sides and the angles opposite, will be equal.

From any point D in AC the base of the isosceles triangle ABC let DE, DF be drawn making the angles CDE, ADF equal to one another; join AE, CF; the triangles AED, CDF are equal.

E

B

Since the angle ADF is equal to the angle EDC, and FADECD, the triangles ECD, FAD are equiangular, and AD : DC :: FD: DE. Also since the angle FDA is equal to EDC, add to each the angle FDE, .. the angle ADE = CDF; hence the sides about the equal angles are reciprocally proportional, and .. (Eucl. vi. 15.) the triangles ADE, FCD are equal.

N

(11.) If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides; these together shall be equal to a perpendicular drawn from either extremity of the base to the opposite side.

Let ABC be an isosceles triangle, from any point D in the base of which, let DE, DF be drawn perpendicular to the sides; and from Blet BG be drawn perpendicular to AC; BG is equal to DE and DF together.

E

B

D

G

F

Since the angle EBD is equal to the angle at C, and the angles at E and F are right angles, the triangles BED, DFC are equiangular, and

.. BD: DC :: DE: DF,

comp. BC: DE+ DF:: DC: DF

But BG being parallel to DF, DC: DF :: BC: BG, whence BC: BG :: BC: DE + DF, and, BG= DE + DF.

(12.) Of all triangles having the same vertical angle, and whose bases pass through a given point, the least is that whose base is bisected in the given point.

Let BAC be the vertical angle of any number of triangles, whose bases pass through a given point P, and let BC be bisected in P: ABC is less than any other triangle ADE.

D

B

F

From C draw CF parallel to AB; then the angle DBP is equal to PCF, and the vertically opposite angles DPB, CPF are equal, and BP = PC,

..the triangle DBP is equal to the triangle PCF, and :. DPB is less than CPE; add to each the trapezium ADPC, and ABC is less than ADE. In the same manner ABC may be proved to be less than any other triangle whose base passes through P.

(13.) If from the angles at the base of a triangle perpendiculars be let fall on a line which bisects the vertical angle; the part of this line intercepted between these perpendiculars will be bisected by a perpendicular from the middle of the base.

From A and B let perpendiculars AD, BG be drawn to the line CD which bisects the angle at C; the part GD will be bisected by a perpendicular EF from E the middle point of the base AB.

A

C

H

B

G

F

E

Produce BG, FE to H and I. Then IE being parallel to HB, and AE= EB, .. (Eucl. vi. 2.) AI= IH. Also since AD is parallel to HG and IF,

DF FG AI : IH,

whence DF FG, and DG is bisected in F.

=

(14.) If from one of the angles at the base of a triangle a line be drawn parallel to the opposite side, and from any point in it lines be drawn making any angles with the sides (produced, if necessary); they will have the same ratio that lines have, which are drawn parallel to them from the other angles, and terminated by the same sides.

From A one of the angles of the triangle ABC let AD be drawn parallel to BC the opposite side; and from any point D in it, let DE, DH be drawn. making any angles with the sides; draw BF, CG parallel to them respectively; DE: DH :: BF: CG.

Since DE is parallel to BF, and DA to BC, the triangles DEA, BFC are equiangular,

.. DE: DA :: BF : BC;

and in a similar manner it may be shewn, that DA: DH:: BC: CG,

.. DE : DH :: BF : CG.

(15.) To bisect a given triangle by a line drawn from one of its angles.

Let ABC be the given triangle, and A the angle, from which the bisecting line is to be drawn. Bisect the opposite side AC in D, and join AD; AD bisects the triangle.

B

D

For the bases BD, DC being equal, (Eucl. i. 38.) the triangles ABD, ADC are also equal.

(16.) To bisect a given triangle by a line drawn from a given point in one of its sides.

Let ABC be the given triangle, and P the given point. Bisect BC in D, join AD, PD; and from A draw AE parallel to PD; join PE; PE bisects the triangle ABC.

Since AE is parallel to PD, the triangle APD is equal to the triangle EPD; from each of them take away the triangle PFD, and AFP = EFD. Also since BD is equal to DC, the triangle ABD is equal to the triangle ADC; parts of which EFD, AFP are equal, .. ABEF is equal to PFDC; whence ABEF and AFP together, or ABEP will be equal to PFDC and FED together, i. e. to PEC; and .. the triangle ABC is bisected by PE.

(17.) To determine a point within a given triangle, from which lines drawn to the several angles, will divide the triangle into three equal parts.

Let ABC be the given triangle; bisect AB, BC, in E, and D; join AD, CE, BF; Fis the point required.

=

B

E

Since BD= DC, the triangle BAD is equal to DAC; and for the same reason the triangle BFD is equal to DFC; .. the triangle BFA is equal to AFC. Again, since BE EA, the triangle BEC is equal to the triangle AEC; parts of which, the triangles BEF, AEF are equal; .. the triangle BFC is equal to AFC; and.. the three BFC, BFA, AFC are equal to one another.

(18.) To trisect a given triangle from a given point within it.

Let ABC be the given triangle, and P the given point within it. Trisect the side BC in D and E; join