Question

In the given figure, AB || DC and AB = DC. Prove that ∆ABD congurent ∆CDB. Hence show that AD || BC.​

Answer 1

Answer:

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

Step-by-step explanation:

Step 1: A polygon having three edges and three vertices is called a triangle. It is one of the fundamental geometric forms. Triangle ABC is the designation for a triangle with vertices A, B, and C. In Euclidean geometry, any three points that are not collinear produce a distinct triangle and a distinct plane.

Step 2: 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)

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