In right ∆ABC the altitude CH to the hypotenuse AB intersects angle bisector AL in point D. Find BC if AD = 8 cm and DH = 4 cm.

check the picture below.

notice that the triangle ADH, since the segment AL is an angle bisector, meaning it cuts the angle A in two equal halves, then the triangle ADH is only using half of A.

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eudora eudora · Answer 2 · 2021-11-19T19:53:33+01:00

Measure of side BC in the figure attached is 24 units.

Given in the question,

m∠CAL = m∠BAL [AL is an angle bisector of ∠CAB]
m∠C = 90°
AD = 8 cm
DH = 4 cm

By applying sine rule in right ΔADH,

[tex]\text{sin}\theta=\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]

[tex]\text{sin}(\angle DAH)=\frac{\text{DH}}{\text{AD}}[/tex]

[tex]=\frac{4}{8}[/tex]

[tex]\angle DAH=\text{sin}^{-1}}(\frac{1}{2} )[/tex]

[tex]=30^\circ[/tex]

Therefore, [tex]\angle CAB=2(\angle DAH)[/tex]

[tex]=2(30^\circ)[/tex]

[tex]=60^\circ[/tex]

By applying tangent rule in right ΔAHD,

[tex]\text{tan}(\angle DAH)=\frac{\text{DH}}{\text{AH}}[/tex]

[tex]\text{tan}(30^\circ)=\frac{4}{AH}[/tex]

[tex]\frac{1}{\sqrt{3}}=\frac{4}{AH}[/tex]

[tex]AH=4\sqrt{3}[/tex]

By applying cosine rule in ΔAHC,

[tex]\text{cos}(\angle CAH)=\frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]

[tex]\text{cos}(60^\circ)=\frac{\text{AH}}{\text{AC}}[/tex]

[tex]\frac{1}{2} =\frac{4\sqrt{3} }{AC}[/tex]

[tex]\text{AC}=8\sqrt{3}[/tex]

By applying tangent rule in right ΔACB,

[tex]\text{tan}(\angle CAB)=\frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]

[tex]\text{tan}(60)^\circ=\frac{BC}{AC}[/tex]

[tex]\sqrt{3}=\frac{BC}{8\sqrt{3}}[/tex]

[tex]BC=(8\sqrt{3})(\sqrt{3})[/tex]

[tex]BC=24[/tex] units

Therefore, measure of side BC is 24 units.

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