1.

In the adjoining figure, AB = AC and BD = DC. Prove that ΔADB ≅ ΔADC and hence show that(i) ∠ADB = ∠ADC = 90o, (ii) ∠BAD = ∠CAD.

Answer»

Given,

AB = AC and BD = DC

To prove,

ΔADB ≅ ΔADC

Proof,

In the right triangles ADB and ADC, we have:

Hypotenuse AB = Hypotenuse AC (given)

BD = DC (given)

AD = AD (common)

∴ ΔADB ≅ ΔADC

By SSS congruence property:

∠ADB = ∠ADC (corresponding parts of the congruent triangles) … (1)

∠ADB and ∠ADC are on the straight line.

∴∠ADB + ∠ADC =180o

∠ADB + ∠ADB = 180o

2 ∠ADB = 180o

∠ADB = 180/2

∠ADB = 90o

From (1):

∠ADB = ∠ADC = 90o

(ii) ∠BAD = ∠CAD (∵ corresponding parts of the congruent triangles)



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