O is any point on the bisector of the acute angle $∠$ XYZ. The line OP $∥$ ZY. Then, $△$ YPO is an Isosceles right-angled triangle.

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Isosceles but not right angled

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Solution

The correct option is B
Isosceles but not right angled

$∠$ POY = $∠$ ZYO (alt. $∠$ s) = $∠$ PYO.

So, PY = PO.

Now, $∠$ XYZ $<$ $90^{\circ}$

$\Rightarrow$ POY $<$ $45^{\circ}$ $\Rightarrow$ $∠$ PYO $<$ $45^{\circ}$

$∴$ $∠$ YPO $>$ $90^{\circ}$

Hence, $Δ$ YPO is an isosceles but not a right angled triangle.

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O is any point on the bisector of the acute angle ∠XYZ. The line OP∥ZY. Then, △YPO is an Isosceles right-angled triangle.

The correct option is B Isosceles but not right angled ∠POY = ∠ZYO (alt. ∠s) = ∠PYO. So, PY = PO. Now, ∠XYZ < 90∘ ⇒POY < 45∘ ⇒∠PYO < 45∘ ∴∠YPO > 90∘ Hence, Δ YPO is an isosceles but not a right angled triangle.

O is any point on the bisector of the acute angle $∠$ XYZ. The line OP $∥$ ZY. Then, $△$ YPO is an Isosceles right-angled triangle.