Question

Proof: In AADC, ZADC= 90° given given Seg DB 1 hypt. AC :: BD2 = Х Theorem of geometric mean :: AB u :: AC = AB + BC = U​

Answer 1

Explanation:

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 )

Now, ∠ADB+∠ADC=180°

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

⇒∠ADB+∠ADB=180° [from(1)]

⇒2∠ADB=180°

⇒∠ADB= ²180°

=90°

∴∠ADB=∠ADC=90° [from (1)]

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