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PROPOSITION VIII. THEOREM.

286. Two triangles which have their sides respectively perpendicular to each other are similar.

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In the triangles EFD and BA C, let EF, FD and ED, be perpendicular respectively to A C, BC and A B. We are to prove A EFD and BA C similar.

Place the A EFD so that its vertex E will fall on A B, and the side EF, 1 to A C, will cut A C at F'.

Draw F'D' to FD, and prolong it to meet BC at H. In the quadrilateral B E D'H, E and H are rt. 4.

..ZB+ZED H= 2 rt. .

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But

LED F+ZED H= 2 rt. 4.

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Now

and

.. LED' F' = ▲ B. ZC+ZHF' C = rt. ≤,

(in a rt. ▲ the sum of the two acute = a rt. ▲);

ZEF D'+ZHF'C= rt. Z.

Ax. 3.

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Ax. 9.

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287. SCHOLIUM. When two triangles have their sides respectively parallel or perpendicular, the parallel sides, or the perpendicular sides, are homologous.

PROPOSITION IX. THEOREM.

288. Lines drawn through the vertex of a triangle divide proportionally the base and its parallel.

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In the triangle ABC let HL be parallel to A C, and let BS and BT be lines drawn through its vertex to the base.

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(two which are mutually equiangular are similar).

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Ex. Show that, if three or more non-parallel straight lines divide two parallels proportionally, they pass through a common point.

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289. If in a right triangle a perpendicular be drawn from the vertex of the right angle to the hypotenuse:

I. It divides the triangle into two right triangles which are similar to the whole triangle, and also to each other.

II. The perpendicular is a mean proportional between the segments of the hypotenuse.

III. Each side of the right triangle is a mean proportional between the hypotenuse and its adjacent segment.

IV. The squares on the two sides of the right triangle have the same ratio as the adjacent segments of the hypote

nuse.

V. The square on the hypotenuse has the same ratio to the square on either side as the hypotenuse has to the segment adjacent to that side.

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In the right triangle ABC, let BF be drawn from the vertex of the right angle B, perpendicular to the hypotenuse A C.

I. We are to prove

the AABF, ABC, and FBC similar.

In the rt. A BA F and BA C,

the acute A is common.

.. the A are similar,

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Now as the rt. A ABF and CBF are both similar to

A B C, by reason of the equality of their ,

they are similar to each other.

II. We are to prove

A F : BF:: BF: FC.

In the similar A ABF and CBF,

A F, the shortest side of the one,
: BF, the shortest side of the other,
:: BF, the medium side of the one,

: FC, the medium side of the other.

III. We are to prove A C : A B :: AB AF.
In the similar ▲ A B C and A B F,

A C, the longest side of the one,
: AB, the longest side of the other,
:: A B, the shortest side of the one,
A F, the shortest side of the other.

Also in the similar ▲ A B C and FBC,

A C, the longest side of the one,
: BC, the longest side of the other,
:: BC, the medium side of the one,
: FC, the medium side of the other.

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In the proportion AC AB::AB: AF,

ABAC XAF,

(the product of the extremes is equal to the product of the means).

and in the proportion AC: BC:: BC: FC,

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Cancel the common factor A C, and we have

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§ 259

(Case III.)

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PROPOSITION XI. THEOREM.

290. If two chords intersect each other in a circle, their segments are reciprocally proportional.

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Let the two chords AB and EF intersect at the

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(two are similar when two of the one are equal to two of the other).

Whence

A O, the medium side of the one,
: EO, the medium side of the other,
:: OF, the shortest side of the one,
: OB, the shortest side of the other.

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Q. E. D.

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