compare AD x AC with AB x AE , for intersecting circles

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compare AD x AC with AB x AE , for intersecting circles

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Source: 2017 Moldova NMO 10.7

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#1 h Mar 17, 2021, 5:42 AM

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Two circles have a common chord $AB$ . Through point $B$ goes a straight line which intersects the circles at points $C$ and $D$ , so that $B$ is between $C$ and $D$ . The tangents to the circles, taken through points $C$ and $D$ , intersect at a point $E$ . Compare $AD \cdot AC$ with $AB \cdot AE$ .

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#2 h Mar 17, 2021, 7:47 AM

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We have $\measuredangle CAD=\measuredangle CAB+\measuredangle BAD=\measuredangle ECD+\measuredangle CDE=\measuredangle CED,$ hence $ACED$ is cyclic.

Now we have $\measuredangle CAB=\measuredangle ECD=\measuredangle EAD$ and $\measuredangle ABC=\measuredangle ABD=\measuredangle ADE$ , therefore $\triangle ABC\sim\triangle ADE$ , hence
$\frac{AB}{AC}=\frac{AD}{AE}\implies AD\cdot AC=AB\cdot AE.$

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