Question

5. In the adjoining figure, AB = AC and BD = DC.
Prove that AADBAADC and hence show that
(1) angle ADB = LADC = 90°. (ii) angle BAD = 2CAD.

TELL ANSWER​

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

=180° ∠ADB + ∠ADB

= 180° 2 ∠ADB

= 180° ∠ADB

= 180/2 ∠ADB

= 90° From (1): ∠ADB = ∠ADC = 90o (ii) ∠BAD = ∠CAD

$❀✿THA᭄NKS★$

Answer 2

Answer: In this we have to use SSS criterion