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Question

In fig. two equal chords $$AB$$ and $$CD$$ of a circle with centre $$O$$, intersect each other at $$E$$. Prove that $$AD = CB$$. 
1235721_7076f0391f8549a091b750028c94f76a.jpg


Solution

From the figure there are two triangles $$ΔAED$$ and $$ΔCEB$$
To prove $$ΔAED$$ and $$ΔCEB$$
From the figure we get$$,$$
$$<AED = <CEB ($$Vertically opposite angles are equal$$)$$
$$<A = < C   ($$since $$<A$$ and $$<C$$ are on the angle subtended on the same arc$$)$$
Similarly $$<D = <B$$
All angle are equal. Therefore $$ΔAED ≅ ΔCEB$$
To prove $$AD = CB$$
Since $$ΔAED ≅ ΔCEB$$
By$$ CPCT ($$Corresponding parts of congruent triangles$$)$$
$$AE = CE, ED = EB$$ and $$AD = CB$$

Mathematics

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