Question
Answer:
:
Properties: 1 Mark
Each proof: 1Mark
In ΔADBandΔADC
AB=AC [Given]
∠ADB=∠ADC=90∘ [Given]
AD=AD [common]
Hence, ΔADB≅ΔADC [By RHS congruence rule…….(1)]
From (1), ∠BAD=∠CAD [Corresponding parts of congruent triangles]
From (1), BD=DC [Corresponding parts of congruent triangles]
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:
Properties: 1 Mark
Each proof: 1Mark
In ΔADBandΔADC
AB=AC [Given]
∠ADB=∠ADC=90∘ [Given]
AD=AD [common]
Hence, ΔADB≅ΔADC [By RHS congruence rule…….(1)]
From (1), ∠BAD=∠CAD [Corresponding parts of congruent triangles]
From (1), BD=DC [Corresponding parts of congruent triangles]
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