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In △ABC\triangle ABC, AB=AC=12AB=AC=12 cm and D is a point on side BC such that AD=8AD=8 cm. If AD is extended to a point E such that ∠ACB=∠AEB\angle ACB=\angle AEB, then the length, in cm, of AE is

Solution

✅ Correct Option: 2

In △ABC\triangle ABC, we have AB=AC=12AB = AC = 12 cm, AD=8AD = 8 cm where DD lies on BCBC, and ADAD is extended to EE such that ∠ACB=∠AEB\angle ACB = \angle AEB.


Since ∠ACB=∠AEB\angle ACB = \angle AEB and both angles subtend the same segment ABAB from the same side, the four points A,B,C,EA, B, C, E are concyclic (they all lie on a common circle).

This means chords AEAE and BCBC intersect at point DD inside the circle. By the Intersecting Chords Theorem:

DA⋅DE=DB⋅DCDA \cdot DE = DB \cdot DC


To find DB⋅DCDB \cdot DC, we apply Stewart's Theorem on △ABC\triangle ABC with cevian ADAD:

AB2⋅DC+AC2⋅BD−AD2⋅BC=BC⋅BD⋅DCAB^2 \cdot DC + AC^2 \cdot BD - AD^2 \cdot BC = BC \cdot BD \cdot DC

Since AB=AC=12AB = AC = 12, we get:

144(DC+BD)−64⋅BC=BC⋅BD⋅DC144(DC + BD) - 64 \cdot BC = BC \cdot BD \cdot DC

Noting that DC+BD=BCDC + BD = BC:

144⋅BC−64⋅BC=BC⋅BD⋅DC144 \cdot BC - 64 \cdot BC = BC \cdot BD \cdot DC

80⋅BC=BC⋅BD⋅DC80 \cdot BC = BC \cdot BD \cdot DC

Dividing both sides by BCBC:

BD⋅DC=80BD \cdot DC = 80


Now applying the Intersecting Chords Theorem:

DA⋅DE=DB⋅DCDA \cdot DE = DB \cdot DC

8⋅DE=808 \cdot DE = 80

DE=10DE = 10 cm


Since EE lies on the extension of ADAD beyond DD:

AE=AD+DEAE = AD + DE

AE=8+10AE = 8 + 10

AE=18AE = 18 cm

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