In isosceles ∆ABC,BC=12,AC=AB=9,find altitude AD
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GIVEN : ∆ABC is an isosceles ∆ in which AB = AC = 13cm and altitude AD = 5 cm.
In ∆ADB, by Pythagoras theorem
AB² = AD² + BD²
13² = 5² + BD²
169 = 25 + BD²
169 − 25 = BD²
BD² = 144
BD = √144
BD = 12 cm
In ∆ ADB & ∆ADC,
∠ADB = ∠ADC [Each 90°]
AB = AC (GIVEN)
AD = AD [Common]
∆ADB ≅ ∆ADC [By RHS criterion]
∴ BD = CD [By c.p.c.t]
BC = BD + CD
BC = 12 + 12
BC = 24 cm
Hence, the length of BC is 24 cm.
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