Question

In Fig. 10.40, it is given that RT=TS, ∠1 =2∠2 and ∠4 = 2∠3. Prove that Δ RBT ≅ Δ SAT.

Answer 1

Answer:

ASA CRITERION

Step-by-step explanation:

RT = RS [given]

$\angle \: TRS= \angle \: TSR \\ (angle \: opposite \: to \: equal \: side \: are \: equal) \\ \\ \angle 1 = \angle 4 \: (vertically \: opposite \: angle) \\ \\ \angle 1 = 2\angle 2 \: (given) \\ \\ \angle 4 = 2\angle 3 \: (given) \\ \\ 2\angle 2 \: = 2\angle 3 \\ (things \: equal \: to \: equal \: are \: equal) \\ \\ \angle 2 \: = \angle 3 \: \\ \\ \angle \: TRS -\angle 2 = \angle \: TSR - \angle 3 \\ \\ \angle TRB = \angle TSA$

Now angle T is common in both the triangles.

RT=RS (given)

Thus by ASA CRITERION OF CONGRUENCY

$\triangle \: RBT \cong \: \triangle \: SAT$

Hope it helps you.

Answer 2

Step-by-step explanation:

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