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Zorluk: ZorRight Triangles and the Pythagorean Theorem

In right triangle ABCABC, the measure of angle CC is 90∘90^\circ. A point DD lies on side ACAC such that BDBD is the angle bisector of angle ABCABC. If BC=28BC = 28 and BD=35BD = 35, what is the length of segment ADAD?

Cevap: 75

Cevap

The length of segment ADAD is 75.
By applying the Pythagorean theorem to right triangle BCDBCD, we find CD=21CD = 21. Then, by the angle bisector theorem, the ratio of ABAB to BCBC equals the ratio of ADAD to CDCD, which gives AB=43ADAB = \frac{4}{3}AD. Using the Pythagorean theorem on right triangle ABCABC, we solve (43AD)2=282+(21+AD)2(\frac{4}{3}AD)^2 = 28^2 + (21 + AD)^2 to find the positive length AD=75AD = 75.

Adım Adım Çözüm

1
Find the length of CDCD using the Pythagorean theorem on right triangle BCDBCD.
CD=21CD = 21
Since angle CC is a right angle, triangle BCDBCD is a right triangle with hypotenuse BDBD and leg BCBC.
2
Apply the angle bisector theorem to express ABAB in terms of ADAD.
AB=43ADAB = \frac{4}{3}AD
The angle bisector theorem states that ADCD=ABBC\frac{AD}{CD} = \frac{AB}{BC}. Substituting BC=28BC = 28 and CD=21CD = 21 gives AD21=AB28\frac{AD}{21} = \frac{AB}{28}.
3
Set up a quadratic equation using the Pythagorean theorem on right triangle ABCABC.
(43AD)2=282+(21+AD)2(\frac{4}{3}AD)^2 = 28^2 + (21 + AD)^2
In right triangle ABCABC, the hypotenuse is ABAB and the legs are BC=28BC = 28 and AC=CD+AD=21+ADAC = CD + AD = 21 + AD.
4
Solve the quadratic equation for ADAD.
AD=75AD = 75
Expanding and simplifying the equation yields AD2−54AD−1575=0AD^2 - 54AD - 1575 = 0, which factors as (AD−75)(AD+21)=0(AD - 75)(AD + 21) = 0. Since length must be positive, AD=75AD = 75.

Anahtar Kavram

Pythagorean Theorem and Angle Bisector Theorem

Alternatif Yöntem

Let θ=∠DBC\theta = \angle DBC. Since BDBD bisects angle BB, ∠ABC=2θ\angle ABC = 2\theta. In right triangle BCDBCD, cos⁡θ=BCBD=2835=45\cos\theta = \frac{BC}{BD} = \frac{28}{35} = \frac{4}{5}. In right triangle ABCABC, cos⁡(2θ)=BCAB=28AB\cos(2\theta) = \frac{BC}{AB} = \frac{28}{AB}. Using the double-angle identity cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1, we get 28AB=2(45)2−1=725\frac{28}{AB} = 2(\frac{4}{5})^2 - 1 = \frac{7}{25}, which gives AB=100AB = 100. Finally, AC=AB2−BC2=1002−282=96AC = \sqrt{AB^2 - BC^2} = \sqrt{100^2 - 28^2} = 96, so AD=AC−CD=96−21=75AD = AC - CD = 96 - 21 = 75.
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Right Triangles and the Pythagorean Theorem

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