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Marks' first lessons in geometry / In two parts. Objectively presented, and designed for the use of primary classes in grammar schools, academies, etc. cover

Marks' first lessons in geometry / In two parts. Objectively presented, and designed for the use of primary classes in grammar schools, academies, etc.

Chapter 137: PROPOSITION XXV. THEOREM. DEMONSTRATION.
By Bernhard Marks
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About This Book

This elementary geometry manual presents a structured course for primary and grammar school pupils, dividing instruction into a development section that introduces points, lines, angles, polygons, triangles, quadrilaterals, circles, and surface measurement using diagrams and guided exercises, and a second section that states axioms and illustrated theorems. Lessons include teacher notes for large classes or instructors with limited formal training, use repetition and review to reinforce concepts, and pose suggestive questions intended to strengthen reasoning alongside perceptive skills so students can continue geometric study independently.

PROPOSITION XXIII. THEOREM.

DEMONSTRATION.

We wish to prove that

A tangent and chord parallel to it intercept equal arcs of the circumference.

Let A B be tangent to the circumference at the point D, and let C F be a chord parallel to the tangent; then will the intercepted arcs C D and D F be equal.

For from the point of contact D, draw the straight line D C.

Because the tangent and chord are parallel, the interior alternate angles A D C and D C F are equal.

But the angle A D C, being formed by the tangent D A and the chord D C, is measured by half the intercepted arc D C;

And the angle D C F, being at the circumference, is measured by half the arc on which it stands, D F:

Then, because the angles are equal, the half arcs which measure them are equal, and the arcs themselves are equal.

PROPOSITION XXIV. THEOREM.

DEMONSTRATION.

We wish to prove that

The angle formed by the intersection of two chords in a circle is measured by half the sum of the intercepted arcs.

Let the chords A B and C D intersect each other in the point E; then will the angle B E D or A E C be measured by half the sum of the arcs A C, B D.

For from the point C draw C F parallel to A B.

Because the chords A B and C F are parallel, the arcs A C, B F, are equal.

Add each of these equals to B D, and we have B D plus A C equal to B D plus B F; that is, the sum of the arcs B D, A C, is equal to the arc F D.

Because the chords A B, C F, are parallel, the opposite exterior and interior angles D E B, D C F, are equal.

But D C F is an angle at the circumference, and is therefore measured by half the arc F D.

Then the equal angle D E B must be measured by half of the arc F D, or its equal B D, plus A C.

PROPOSITION XXV. THEOREM.

DEMONSTRATION.

We wish to prove that

The angle formed by two secants meeting without a circle is measured by half the difference of the intercepted arcs.

Let the secants A B, A C, intersect the circumference in the points D and E; then will the angle B A C be measured by half the difference between the arcs B C and D E.

For from the point D draw the chord D F parallel to E C.

Because A C and D F are parallel, the opposite exterior and interior angles B D F and B A C are equal.

Because the chords D F, E C, are parallel, the arcs D E and F C are equal.

If from the arc B C we take the arc D E, or its equal F C, we shall have left the arc B F;

But the angle B D F, being at the circumference, is measured by half the arc B F:

Then the equal of B D F, or B A C, must be measured by half the arc B F, or half the difference between the intercepted arcs B C and D E.