Right angle abc,bd perpendicular ac,if bd=8cm,ad=4cm,find cd and ab
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AB=√(AD²+BD²)
= 4√5cm
in Δ BCD
BD⏊CD
∠DBC+∠ DCB = 90°.......(1)
In Δ ABC AB⏊BC
∠BAC+∠ACB= 90°.........(2)
equating (1) &(2)
∠BAC+∠ACB=∠DBC+∠ DCB
∠BAC= ∠ DBC [∵ ∠ACB & ∠ DCB are same angles]......(3)
inΔ ABD & Δ DBC
∠BAC= ∠ DBC[ using (3)
∠BDC= ∠ADB [ ∵ BD⏊AC]
BD is common
Δ ABD≅Δ DBC[AAS]
AB=BC= 4√5 cm [ CPCT]
= 4√5cm
in Δ BCD
BD⏊CD
∠DBC+∠ DCB = 90°.......(1)
In Δ ABC AB⏊BC
∠BAC+∠ACB= 90°.........(2)
equating (1) &(2)
∠BAC+∠ACB=∠DBC+∠ DCB
∠BAC= ∠ DBC [∵ ∠ACB & ∠ DCB are same angles]......(3)
inΔ ABD & Δ DBC
∠BAC= ∠ DBC[ using (3)
∠BDC= ∠ADB [ ∵ BD⏊AC]
BD is common
Δ ABD≅Δ DBC[AAS]
AB=BC= 4√5 cm [ CPCT]
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