In the following figure , OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q. Prove that : (i) $Δ O P A ≅ Δ O Q C$

Open in App

Solution

In $△ O P A$ and $△ O Q C$
$O P = O Q$ (radii of same circle)
$∠ A O P = ∠ C O Q (b o t h 90^{o})$
$O A = O C$ (sides of the square)
By side-Angle-Side criterian of congruence,
$∴ △ O P A ≅ △ O Q C (b y S A S)$

Suggest Corrections

Similar questions

Q. In given figure, O is the centre of a circle and ∠ABC = 45°. Prove that OA ⊥ OC.

Q. In the give figure, OPQR is square. A circle drawn with centre O cuts the square in X and Y. Prove that OX = QY.

Q. In the adjoining figure, OPQR is a square. A circle drawn with centre O cuts the square at X and Y. Prove that QX = QY.

Q. In circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with OA as diameter to cut AD at C. Prove that BD = 2 OC.

Two circle with centres O and O' are drawn to intersect each other at points A and B.

Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O' at A. Prove that OA bisects $∠ B A C$

Join BYJU'S Learning Program

In the following figure , OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q. Prove that : (i) ΔOPA≅ΔOQC

In △OPA and △OQCOP=OQ (radii of same circle)∠AOP=∠COQ (both 90o)OA=OC (sides of the square)By side-Angle-Side criterian of congruence,∴△OPA≅△OQC (by SAS)

In the following figure , OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q. Prove that : (i) $Δ O P A ≅ Δ O Q C$

In $△ O P A$ and $△ O Q C$
$O P = O Q$ (radii of same circle)
$∠ A O P = ∠ C O Q (b o t h 90^{o})$
$O A = O C$ (sides of the square)
By side-Angle-Side criterian of congruence,
$∴ △ O P A ≅ △ O Q C (b y S A S)$