In right-angled triangles Delta OXM and Delta OYP, we have right angles OMX = OPY = 90^°, equal hypotenuses OX = OY, and equal legs OM = OP. By the RHS (Right angle-Hypotenuse-Side) congruence criterion, Delta OXM Delta OYP.

In right-angled triangles Delta OXM and Delta OYP, we have right angles OMX = OPY = 90^°, equal hypotenuses OX = OY, and equal legs OM = OP. By the RHS (Right angle-Hypotenuse-Side) congruence criterion, Delta OXM Delta OYP.

Fig. 13.36 shows a circle with center $O$ . A line segment $BC$ passes through $O$ as a diameter. A chord $AB$ is drawn, and a vertical segment $AC$ connects $A$ to $C$ . A line segment $OD$ is drawn from the center $O$ to chord $AB$ , intersecting $AB$ at point $D$ . Fig. 13.37 shows a square $OMNP$ with vertices labeled clockwise/counterclockwise such that $O$ is at the bottom-right, $P$ at the bottom-left, $N$ at the top-left, and $M$ at the top-right. A circle with center at vertex $O$ intersects the side $MN$ at point $X$ and the side $PN$ at point $Y$ . 12. In Fig. 13.37, $OMNP$ is a square. A circle drawn with centre $O$ cuts the square in $X$ and $Y$ . Prove that : $\Delta OXM \cong \Delta OYP$ . Hence prove that $NX = NY$ .

Answer: $\Delta OXM \cong \Delta OYP$ by the RHS criterion, which implies $XM = YP$ . Consequently, $NX = MN - XM = PN - YP = NY$ .

Step-by-step solution

Step 1: Identify given properties of square and circle

Since $OMNP$ is a square, all of its four sides are equal, giving $OM = OP$ . Furthermore, each interior angle of a square measures $90^\circ$ , so $\angle OMX = \angle OPY = 90^\circ$ .

Step 2: Relate circle radii to the hypotenuses

The points $X$ and $Y$ both lie on the circle centered at $O$ . Therefore, the segments connecting center $O$ to points $X$ and $Y$ are radii of the circle, making $OX = OY$ .

Step 3: Apply RHS congruence criterion

In right-angled triangles $\Delta OXM$ and $\Delta OYP$ , we have right angles $\angle OMX = \angle OPY = 90^\circ$ , equal hypotenuses $OX = OY$ , and equal legs $OM = OP$ . By the RHS (Right angle-Hypotenuse-Side) congruence criterion, $\Delta OXM \cong \Delta OYP$ .

Step 4: Deduce corresponding parts

Since corresponding parts of congruent triangles are congruent (CPCTC), we obtain $XM = YP$ .

Step 5: Prove NX = NY using side subtraction

Since $OMNP$ is a square, the entire side lengths are equal, so $MN = PN$ . Subtracting the equal segments $XM = YP$ from both sides gives $MN - XM = PN - YP$ , which simplifies directly to $NX = NY$ .

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