Challenging Problems in Geometry by Alfred S. Posamentier Charles T. Salkind

Author:Alfred S. Posamentier, Charles T. Salkind
Language: eng
Format: epub, pdf
Publisher: Dover Publications

Both and are complementary to an angle of measure α; therefore, they are congruent. Thus, ΔABE is isosceles, and, AB = BE = x (#5).

4-39If equilateral ΔABC is inscribed in a circle, and a point P is chosen on minor arc , prove that PB = PA + PC (Fig. S4-39).

Choose a point Q on BP such that PQ = QC.

Since ΔABC is equilateral, .

Therefore, .

Since in ΔPQC, PQ = QC, and , ΔQPC is equilateral.

PC = QC and as both are equal in measure to .

Thus, (S.A.A.), and BQ = AP. Since BQ + QP = BP, by substitution, AP + PC = PB.

4-40From point A, tangents are drawn to circle O, meeting the circle at B and C. Chord secant ADE, as in Fig. S4-40. Prove that FC bisects DE.

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