In below shown figure- PS - SQ - PT - TR and - PST - P RQ- Then - PQR is a - an - triangle-A- EquilateralB- IsoscelesC- ScaleneD- Right angled

The correct option is B IsoscelesIt is given that PS/SQ nbsp;= PT/TR So, ST||QR nbsp; nbsp; nbsp;[By converse of Basic Proportionality Theorem] ∴, ∠ ...

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Byju's Answer

Standard X

Mathematics

Converse of Thales Theorem

In below show...

Question

In below shown figure, $\frac{P S}{S Q} = \frac{P T}{T R}$ and $∠ P S T = ∠ P R Q$ . Then $Δ P Q R$ is a/an __triangle.

Equilateral

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Isosceles

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Scalene

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Right angled

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Solution

The correct option is B Isosceles
It is given that

$\frac{P S}{S Q} = \frac{P T}{T R}$

So, $S T | | Q R$ [By converse of Basic Proportionality Theorem]

$∴, ∠ P S T = ∠ P Q R$ (Corresponding Angles)

Also, it is given that

$∠ P S T = ∠ P R Q$

So, $∠ P R Q = ∠ P Q R$

Therefore, $P Q = P R$ (Sides opposite the equal angles)

i.e., $Δ P Q R$ is an isosceles triangle.

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Similar questions

Q. In the figure shown below , if

\frac{P S}{S Q}

\frac{P T}{T R}

and

∠

PST=

∠

PRQ, then

Δ

PQR is a/an triangle.

In below shown figure, $\frac{P S}{S Q}$ = $\frac{P T}{T R}$ and $∠$ PST= $∠$ PRQ. Then $Δ$ PQR is an isosceles Triangle.

In the given figure, $\frac{P S}{S Q}$ = $\frac{P T}{T R}$ and $∠$ PST= $∠$ PRQ, then $Δ$ PQR is _______triangle.

In the figure shown below , if $\frac{P S}{S Q}$ = $\frac{P T}{T R}$ and $∠$ PST= $∠$ PRQ, then $Δ$ PQR is _______ triangle.

In below shown figure, $\frac{P S}{S Q} = \frac{P T}{T R}$ and $∠ P S T = ∠ P R Q$ . Then $Δ$ PQR is an __ triangle.

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In below shown figure, PSSQ=PTTR and∠PST=∠PRQ. Then ΔPQR is a/an __ ____triangle.

In below shown figure, $\frac{P S}{S Q} = \frac{P T}{T R}$ and $∠ P S T = ∠ P R Q$ . Then $Δ P Q R$ is a/an __triangle.