Question

In the adjoining figure, AB=AC, BD=DC. Prove ∆ABD=∆ACD and show that
(a) ADB= ADC=90°
​

Answer 1

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)

Answer 2

Answer:

Given

AB=AC and BD=DC

To prove:

△ABD≅△ADC

Proof:

In △ADB and △ADC, we have

AB=AC (given)

BD=DC (given)

AD=AD (common)

∴ △ADB≅△ADC [By SSS congruence property]

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

Now, ∠ADB+∠ADC=180 [∵∠ADB and ∠ADC are on the straight line]

⇒∠ADB+∠ADB=180

[from(1)]

⇒2∠ADB=180

⇒∠ADB=(180)/2

=90

∴∠ADB=∠ADC=90

[from (1)]

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

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