Question

5. In the adjoining figure, AB = AC and BD = DC.
Prove that AADB = AADC and hence show that
(i) ZADB = ZADC = 90°, (ii) ZBAD = ZCAD.

Answer 1

Step-by-step explanation:

Solution

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Given

AB=AC and BD=DC

To prove:

△≅△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

o

[∵∠ADB and ∠ADC are on the straight line]

⇒∠ADB+∠ADB=180

o

[from(1)]

⇒2∠ADB=180

o

⇒∠ADB=

2

180

o

=90

o

∴∠ADB=∠ADC=90

o

[from (1)]

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