(1)

Consider the similar triangles PAG and PC B

Now, examine similar triangles QCA and QBD for ther ratio of CQ and BQ.

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(2)

Consider the similar triangles GFB and PCB.

Likewise, consider similar triangles AED and ACQ.

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(3)

Ceva's theorem (acutally the converse) would indicate that if the product of the ratios determined the the same order by the Cevians, BP, AP, and CK is equal to 1 then the three Cevians are concurrent. Than is, we want to evaluate

From (1)

and .

Since CK is the geometric mean of KB and AK, we can arrive at

and so the product is 1. Therefore the three Cevians are concurrent. Since AD and BG meet at H, therefore H is on CK.

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(4)

Use triangle CKB with AD as the transversal. We are working toward finding a relation involving CH.

By Menelaus we know that

Using the similarities of the various right triangles,

and from (1),

Therefore

The symmetry of this relation would argue that the same result would have been obtained in using right triangle AKC with BG as the transversal.

We need to find a value for HK and it will be useful to do so in terms of AB and CK.

Since and substitute to get

This expression can now be used to substitue in the previous expression.

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