Question

answer please. Show angle TQS = ANGLE QRT+ QSR

Answer 1

Since,

∆PTR ≅ ∆QTS,

By CPCT

PT = QT

TR = TS

PR = QS

∠PTR = ∠QTS

∠TRP = ∠TSQ

∠RPT = ∠SQT

Let,

a = ∠TRQ

b = ∠QSR

x = ∠PTQ

Thus,

In ∆TPQ

∠TPQ = ∠TQP = 90 - $\frac{x}{2}$                     [Angle Sum and Isoceles Triangle]

Similarly,

In ∆TRS,

∠TRS = ∠TSR = 90 - $\frac{x}{2}$                     [Angle Sum and Isoceles Triangle]

Now,

90 - $\frac{x}{2}$ + 90 - [Angle sum in quadrilateral]

a = 90 - $\frac{x}{2}$ - y

Also,

90 - $\frac{x}{2}$ - y + b = 90 - [Angle S]

y = b

∠TQS = 90 - $\frac{x}{2}$

= 90 - $\frac{x}{2}$ - y + y        [Adding and subtracting 'y']

= ∠QRT + ∠QSR              [∠QRT = a, ∠QSR = b]

Therefore Proved

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