Question

. In the adjoining figure, AB AC and BD = DC
Prove that ADB ADC and hence show that
(1) ADB ADC-90 degree (2) BAD - CAD.​

Answer 1

Answer:

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° = 90°

2

∴∠ADB=∠ADC=90° [from (1)]

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